复杂数字的权力和根源

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Manually calculating (simplifying) a statement such as: $\begin{align*}(14 - 17i)^5\end{align*}$ or $\begin{align*}\sqrt[4]{(3 - 2i)}\end{align*}$ in present (rectangular) form would be a very intensive process at best.
::以目前(矩形)形式手工计算(简化)以下说明(14-17i)5 或4(3-2i),充其量是一个非常密集的过程。

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Fortunately you will learn in this lesson that there is an alternative: De Moivre's theorem . De Moivre's theorem is really the only practical method for finding the powers or roots of a complex number , but there is a catch...
::幸运的是,你会从这个教训中了解到,有别的办法:德莫伊夫雷的理论。德莫伊夫雷的理论其实是找到复杂数字的力量或根源的唯一实用方法,但有渔获...

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What must be done to a complex number before De Moivre's theorem can be utilized?
::在利用德莫伊夫雷的理论之前,必须对一个复杂数字做些什么?

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Powers and Roots of Complex Numbers
::复杂数字的权力和根源

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Powers of Complex Numbers
::复杂数字的功率

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How do we raise a complex number to a power? Let’s start with an example:
::我们如何将一个复杂的数字提升到一个强国?

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$\begin{align*}(-4 - 4i)^3=(-4 - 4i)\cdot (-4 - 4i)\cdot (-4 - 4i)\end{align*}$
:--4-4i)3=(-4-4-4i)__(-4-4-4i)__(-4-4-4i)__(-4-4-4i)__(-4-4-4-4i)

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In rectangular form , this can get very complex. What about in r cis θ form?
::在矩形形式中,这可能会变得非常复杂。如果以 rcis 的形式呢?

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$\begin{align*}(-4 - 4i) = 4\sqrt{2}\ cis \left ( \frac{5\pi}{4} \right )\end{align*}$
:-4-4i)=4______________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

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So the problem becomes
::所以问题就变成

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$\begin{align*}4\sqrt{2} \ \mbox{cis} \left ( \frac{5\pi}{4} \right ) \cdot \ 4\sqrt{2} \ \mbox{cis}\left ( \frac{5\pi}{4} \right ) \cdot \ 4\sqrt{2} \ \mbox{cis}\left ( \frac{5\pi}{4} \right )\end{align*}$
::4_________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

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and using our multiplication rule from the previous section,
::并使用上一节中的乘法规则,

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$\begin{align*}(-4 - 4i)^3 = (4\sqrt{2})^3 \ \mbox{cis}\left ( \frac{15\pi}{4} \right )\end{align*}$
:-4-4i)3=(-42)3 Cis(15°4)

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Notice, ( a + bi ) ³ = r ³ cis 3 θ
::通知, (a + bi)3 = r3cis 3 ______________________________________________________________________________________________________________________________________________________________________________________

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In words: Raise the r -value to the same degree as the complex number is raised and then multiply that by cis of the angle multiplied by the number of the degree.
::意思是:将r值提高至与复杂数提高的相同程度,然后通过角的精度乘以度数,再乘以角的精度乘以度数。

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Reflecting on the example above, we can identify De Moivre's Theorem :
::从上述例子来看,我们可以确定德莫伊夫雷的理论:

Let z = r (cos θ + i sin θ ) be a complex number in rcisθ form. If n is a positive integer, z n is z n = r n (cos( nθ ) + i sin( nθ ))

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It should be clear that the polar form provides a much faster result for raising a complex number to a power than doing the problem in rectangular form.
::应当清楚的是,极地形式比以矩形形式处理问题,为将一个复杂数字提高到一个力量提供了更快的结果。

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Roots of Complex Numbers
::复杂数字的根根

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You probably noticed long ago that when an new operation is presented in mathematics, the inverse operation often follows. That is generally because the inverse operation is often procedurally similar, and it makes good sense to learn both at the same time.
::很久以前,你可能就注意到,当新的操作在数学中出现时,反向操作往往随之而来。 一般来说,反向操作在程序上往往相似,同时学习两者是有道理的。

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This is no exception:
::这并不例外:

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The inverse operation of finding a power for a number is to find a root of the same number.
::查找数字功率的反向操作是找到相同数字的根。

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1. Recall from algebra that any root can be written as x ^{1/ n}
   ::从代数中提醒提醒代数, 任何根可以写为 x1/ n
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2. Given that the formula for De Moivre’s theorem also works for fractional powers, the same formula can be used for finding roots:
   ::鉴于德莫伊夫雷定理的公式也适用于分权,同一公式可用于寻找根源:

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$\begin{align*}z^{1/n} = (a + bi)^{1/n} = r^{1/n} cis \left ( \frac{\theta}{n} \right )\end{align*}$
::z1/n=( a+bi) 1/n=r1/ncis( n)

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Examples
::实例

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Example 1
::例1

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Earlier, you were asked what should be done to a complex number before you can use De Moivre's theorem on it.
::早些时候,有人问你,在使用德莫伊夫雷的定理之前,应该如何处理一个复杂数字。

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A complex number operation written in rectangular form, such as: $\begin{align*}(13 - 4i)^3\end{align*}$ must be converted to polar form before utilizing De Moivre's theorem.
::以矩形形式写成的复合数操作,例如: (13-4i)3 在使用De Moivre的定理之前必须转换成极形。

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Example 2
::例2

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Find the value of $\begin{align*}(1 + \sqrt{3}i)^4\end{align*}$ .
::查找值(13i) 4。

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$\begin{align*}r = \sqrt{(1)^2 + (\sqrt{3})^2} = 2\end{align*}$
::r(1)2+(%3)2=2

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$\begin{align*}\mbox{tan} \ \theta_{ref} = \frac{\sqrt{3}}{1},\end{align*}$
::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}我... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}我... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}我... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}

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and θ is in the 1 ^st quadrant, so
::和在第一象限,所以

$\begin{align*}\theta = \frac{\pi}{3}\end{align*}$

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Using our equation from above:
::利用我们从上方的方程式:

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$\begin{align*}z^4 = r^4\ cis\ 4\theta\end{align*}$
::z4=r4 cis 4

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$\begin{align*}z^4 = (2)^4\ cis\ 4\frac{\pi}{3}\end{align*}$
::z4=(2)4 cis 43

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Expanding cis form:
::展开 CIS 窗体 :

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$\begin{align*}z^4 = 16\left ( \mbox{cos} \left ( \frac{4\pi}{3} \right ) + i \ \mbox{sin}\left ( \frac{4\pi}{3} \right ) \right )\end{align*}$
::z4=16(cos(4)3+i sin (4)3))

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$\begin{align*}= 16\left ( (-0.5) - 0.866i \right )\end{align*}$
::=16((-0.5)-0.866i)

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Finally we have
::最后,我们

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z ⁴ = -8 - 13.856 i
::z4 = - 8 - 13.856i

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Example 3
::例3

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Find $\begin{align*}\sqrt{1 + i}\end{align*}$ .
::查找% 1+i 。

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First, rewriting in exponential form: (1 + i) ^½
::首先,以指数形式重写(1+1)1⁄2

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And now in polar form:
::现在以极形的形式:

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$\begin{align*}\sqrt{1 + i} = \left (\sqrt{2}\ cis \left (\frac{\pi}{4} \right ) \right )^{1/2}\end{align*}$
::1+1=(%2cis(%4))1/2

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Expanding cis form,
::扩展 CIS 形状,

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$\begin{align*}= \left (\sqrt{2} \left ( \mbox{cos} \left ( \frac{\pi}{4} \right ) + i \ \mbox{sin} \left ( \frac{\pi}{4} \right ) \right ) \right )^{1/2}\end{align*}$
::= = (%2(cos)+i sin (%4))1/2)

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Using the formula:
::使用公式 :

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$\begin{align*}=(2^{1/2})^{1/2} \left ( \mbox{cos} \left ( \frac{1}{2} \cdot \frac{\pi}{4} \right ) + i \ \mbox{sin} \left ( \frac{1}{2} \cdot \frac{\pi}{4} \right ) \right )\end{align*}$
::=(21/2)1/2(cos(12)+i sin (12)4))

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$\begin{align*}= 2^{1/4} \left ( \mbox{cos} \left ( \frac{\pi}{8} \right ) + i \ \mbox{sin} \left ( \frac{\pi}{8} \right ) \right )\end{align*}$
::=21/4(cos(8)+i sin(8))

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In decimal form, we get
::以小数小数的形式,我们得到

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=1.189( 0.924 + 0.383 i )
::=1.189(0.924+0.383i)

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=1.099 + 0.455i
::=1.099 + 0.4555i

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To check, we will multiply the result by itself in rectangular form:
::为了检查,我们将以矩形形式将结果本身乘以:

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$\begin{align*}(1.099 + 0.455i)\ \cdot \ (1.099 + 0.455\mbox{i}) = 1.099^2 + 1.099(0.455i) + 1.099(0.455i) \ +\end{align*}$ $\begin{align*}(0.455i)^2\end{align*}$
:1.099+0.0.455i) (1.099+0.0.455i)=1.0992+1.099(0.455i)+1.099(0.455i)+(0.455i)2

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$\begin{align*}= 1.208 + 0.500i + 0.500i + 0.208i^2\end{align*}$
::=1.208+0.500i+0.500i+0.208i2

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$\begin{align*}= 1.208 + i - 0.208 \ \mbox{or}\end{align*}$
::=1.208+i-0.208或

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$\begin{align*}= 1 + i\end{align*}$
::=1+i =1+i

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Example 4
::例4

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Find the value of $\begin{align*}x : x^3 = \left (1 - \sqrt{3}i \right )\end{align*}$ .
::查找 x: x3 = (1\\\\ 3i) 的值 。

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First we put $\begin{align*}1 - \sqrt{3}i\end{align*}$ in polar form.
::首先,我们把13i 以极形。

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Use $\begin{align*}x = 1, \ y = -\sqrt{3}\end{align*}$ to obtain $\begin{align*}r = 2, \ \theta = \frac{5\pi}{3}\end{align*}$
::使用 x= 1, y3 获取 r= 2, 53

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let $\begin{align*}z = \left ( 1 - \sqrt{3}i \right )\end{align*}$ in rectangular form
::let z=( 1% 3i) 矩形

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$\begin{align*}z = 2\ \mbox{cis}\ \left ( \frac{5\pi}{3} \right )\end{align*}$ in polar form
::z=2 cis (5°3) 极形

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$\begin{align*}x = \left (1 - \sqrt{3}i \right )^{1/3}\end{align*}$
::x=(13i)1/3

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$\begin{align*}x = \left [2cis \left ( \frac{5\pi}{3} \right ) \right ]^{1/3}\end{align*}$
::x=[2Cis( 5_ 3)]1/3

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Use De Moivre’s theorem to find the first solution:
::使用德莫伊夫雷的理论来找到第一个解决方案:

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$\begin{align*}x_1 = 2^{1/3} cis \left ( \frac{5\pi /3}{3} \right )\end{align*}$ or $\begin{align*}2^{1/3} cis \left ( \frac{5\pi}{9} \right )\end{align*}$
::x1=21/3Cis( 5%/ 33) 或 21/3Cis( 5% 9)

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Leave answer in cis form to find the remaining solutions:
::保留答题的精密格式, 以找到其余的解决方案 :

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n = 3 which means that the 3 solutions are $\begin{align*}\frac{2\pi}{3}\end{align*}$ radians apart or
::n=3 = 3,这意味着3个溶液是2×3弧度分开的或

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$\begin{align*}x_2 = 2^{1/3} cis \left ( \frac{5\pi}{9} + \frac{2\pi}{3} \right )\end{align*}$ and $\begin{align*}x_3 = 2^{1/3} cis \left ( \frac{5\pi}{9} + \frac{2\pi}{3} + \frac{2\pi}{3} \right )\end{align*}$
::x2=21/3cis( 5=9+23) 和 x3=21/3cis( 5=9+23+23)

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NOTE: It is not necessary to add $\begin{align*}\frac{2\pi}{3}\end{align*}$ again. Adding $\begin{align*}\frac{2\pi}{3}\end{align*}$ three times equals 2 π . That would result in rotating around a full circle and to start where it all began- that is the first solution.
::注意: 没有必要再增加 2 3。 添加 2 3 乘以 2 3 等于 2 3 。 这将导致整个圆圈旋转, 并且从一切开始的地方开始, 这是第一个解决办法 。

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The three solutions are:
::三种解决办法是:

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$\begin{align*}x_1 = 2^{1/3} cis \left (\frac{5\pi}{9} \right )\end{align*}$
::x1=21/3Cis( 5°9)

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$\begin{align*}x_2 = 2^{1/3} cis \left (\frac{11\pi}{9} \right )\end{align*}$
::x2=21/3Cis( 11_ 9)

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$\begin{align*}x_3 = 2^{1/3} cis \left (\frac{17\pi}{9} \right )\end{align*}$
::x3=21/3Cis(17_9)

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Each of these solutions, when graphed will be $\begin{align*}\frac{2\pi}{3}\end{align*}$ apart.
::这些解决方案中的每一种,当图表显示为 2+3 分开时 。

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Check any one of these solutions to see if the results are confirmed.
::检查其中任何一种解决方案, 以确定结果是否得到确认 。

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Checking the second solution:
::正在检查第二个解决方案 :

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$\begin{align*}x_2 = 2^{1/3} cis \left (\frac{11\pi}{9} \right )\end{align*}$
::x2=21/3Cis( 11_ 9)

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$\begin{align*}= 1.260 \left [ \mbox{cos} \left (\frac{11\pi}{9} \right ) + i \ \mbox{sin} \left (\frac{11\pi}{9} \right ) \right ]\end{align*}$
::=1.260[cos(11)9+i sin(11)9]]

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$\begin{align*}= 1.260[-0.766 - 0.643i]\end{align*}$
::=1.260[- 0.766-0.643i]

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$\begin{align*}= -0.965 - 0.810i\end{align*}$
::0.965-0.810i

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Does (-0.965 – 0.810 i ) ³ or (-0.965 – 0.810 i ) (-0.965 – 0.810 i ) (-0.965 – 0.810 i )
:-0.965 - 0.810i)3 或 (-0.965 - 0.810i) (-0.965 - 0.810i) (-0.965 - 0.810i) (-0.965 - 0.810i)

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$\begin{align*}= \left (1-\sqrt{3}i\right )?\end{align*}$
::=(13i)?

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Example 5
::例5

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What are the two square roots of i ?
::我的两个平方根是什么?

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Let $\begin{align*}z = \sqrt{0 + i}\end{align*}$ .
::让我们z0+i。

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$\begin{align*}r = 1, \ \theta = \pi/2\end{align*}$ or $\begin{align*}z = \left [1 \times \ cis \frac{\pi}{2} \right ]^{1/2}\end{align*}$ Utilizing De Moivre’s theorem:
::r=1, /2 或z=[1x cis2]1/2 利用德莫伊夫雷的定理:

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$\begin{align*}z_1 = \left [1 \times \ cis \frac{\pi}{4} \right ]\end{align*}$ or $\begin{align*}z_2 = \left [1 \times \ cis \frac{5\pi}{4} \right ]\end{align*}$
::z1=[1xcis%4]或z2=[1xcis5%4]

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$\begin{align*}z_1 = 1\left ( \mbox{cos}\frac{\pi}{4} + i \ \mbox{sin}\frac{\pi}{4} \right )\end{align*}$ or $\begin{align*}z_2 = 1 \left (\mbox{cos} \frac{5\pi}{4} + i \ \mbox{sin} \frac{5\pi}{4} \right )\end{align*}$
::z1=1(cos4+i sin_4)或z2=1(cos54+i sin5_4)

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$\begin{align*}z_1 = 0.707 + 0.707i\end{align*}$ or $\begin{align*}z_2 = -0.707 - 0.707i\end{align*}$
::z1=0.707+0.707i或z20.707-0.707i

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Check for z ₁ solution: (0.707 + 0.707 i ) ² = i?
::检查 z1 溶液 0. 707 + 0. 707i) 2 = i?

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0.500 + 0.500 i + 0.500 i + 0.500 i ² = 0.500 + i + 0.500(-1) or i
::0500+0.500i+0.500i+0.500i+0.500i+0.500i2=0.500+i+0.500(-1)或i

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Example 6
::例6

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Calculate $\begin{align*}\sqrt[4]{(1 + 0i)}\end{align*}$ . What are the four fourth roots of 1?
::计算 4( 1+0i) 。 1 的四根基是什么 ?

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Let z = 1 or z = 1 + 0 i . Then the problem becomes find z ^1/4 = (1 + 0 i ) ^1/4.
::Let z = 1 或 z = 1 + 0i. 然后问题会变成 z1/4 = 1 + 0i) 1/4 。

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Since $\begin{align*}r = 1 \ \theta = 0, \ z^{1/4} = [1 \times cis \ 0]^{1/4}\end{align*}$ with $\begin{align*}z_1 = 1^{1/4} \left (\mbox{cos}\ \frac{0}{4} + i \ \mbox{sin}\ \frac{0}{4} \right )\end{align*}$ or $\begin{align*}1(1 + 0)\end{align*}$ or $\begin{align*}1\end{align*}$
::由于 r= 1 0, z1/4 = [1xcis 00] 1/4 和 z1= 11/4 (cos 04+i sin 04 或 1 1+0) 或 1

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That root is not a surprise. Now use De Moivre’s to find the other roots:
::这根根并不令人惊讶。 现在使用德莫伊夫雷(De Moivre)找到另一根根:

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$\begin{align*}z_2 = 1^{1/4} \left [\mbox{cos}\left (0 + \frac{\pi}{2}\right ) + i \ \mbox{sin}\left (0 + \frac{\pi}{2} \right ) \right ]\end{align*}$ Since there are 4 roots, dividing 2π by 4 yields 0.5π
::z2=11/4[cos(02)+i sin(02) 由于有4根, 2除以 4

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or 0 + i or just i $\begin{align*}z_3 = 1^{1/4} \left [\mbox{cos}\left (0 + \frac{2\pi}{2}\right ) + i \ \mbox{sin}\left (0 + \frac{2\pi}{2} \right ) \right ]\end{align*}$ which yields z ₃ = -1
::或 0 + i 或 just i z3=11/4 [cos( 0+22)+i sin( 0+22) , 产生 z3 = -1

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Finally,  $\begin{align*}z_4 = 1^{1/4} \left [\mbox{cos}\left (0 + \frac{3\pi}{2}\right ) + i \ \mbox{sin}\left (0 + \frac{3\pi}{2} \right ) \right ]\end{align*}$ or $\begin{align*}z_4 = -i\end{align*}$
::最后,z4=11/4[cos(0+32)+i sin(0+32)或z4i

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The four fourth roots of 1 are 1, i, -1 and -i.
::1的四根根是1,i,-1和i。

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Example 7
::例7

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Calculate $\begin{align*}(\sqrt{3} + i)^7\end{align*}$ .
::计算 (% 3+i) 7。

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To calculate $\begin{align*}(\sqrt{3} + i)^7\end{align*}$ start by converting to $\begin{align*}rcis\end{align*}$ form.
::要计算 (% 3+i) 7, 请从转换为 rcis 窗体开始 。

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First, find $\begin{align*}r\end{align*}$ . Recall $\begin{align*}r = \sqrt{\sqrt{3}^2 + 1^2}\end{align*}$ .
::首先,找到r. recall r. recall r. 32+12。

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$\begin{align*}r = \sqrt{3 + 1}\end{align*}$
::r 3+1

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$\begin{align*}r = 2\end{align*}$
::r=2

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If $\begin{align*}cos \theta = \frac{\sqrt{3}}{2}\end{align*}$ and $\begin{align*}sin \theta = \frac{1}{2}\end{align*}$ then $\begin{align*}\theta = 30^o\end{align*}$ and is in quadrant I. Now that we have trigonometric form , the rest is easy:
::如果cos32和sin12,那么30o, 位于象限I。 现在我们有了三角形, 其余的很容易:

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$\begin{align*}(\sqrt{3} + i)^7 = [2 (cos 30^o + i sin 30^o)]^7\end{align*}$ ..... Write the original problem in $\begin{align*}r cis\end{align*}$ form
:cos30o+isin30o) 7...。 以 rcis 格式写原问题 。

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$\begin{align*}2^7[ (cos (7 \cdot 30^o) + i sin(7 \cdot 30^o)]\end{align*}$ ..... De Moivre's theorem
::27[(cos( 7)30o)+isin( 7)30o]...。 De Moivre 的定理

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$\begin{align*}128 [-\frac{\sqrt{3}}{2} + \frac{-1}{2}i]\end{align*}$ ..... Simplify
::128 [332112i] .... 简化

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$\begin{align*}(\sqrt{3} + i)^7 = -64\sqrt{3} - 64i\end{align*}$ ..... Simplify again
:% 3+i) 7 643- 64i. 再次简化

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$\begin{align*}\therefore (\sqrt{3} + i)^7 = -64\sqrt{3} - 64i\end{align*}$
::3+i)7643-64i

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Review
::回顾

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Perform the indicated operation on these complex numbers:
::以这些复杂数字执行指定的操作 :

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1. Divide: $\begin{align*}\frac{2 + 3i}{1 - i}\end{align*}$
   ::除数: 2+3i1-i
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2. Multiply: $\begin{align*}(-6 - i)(-6 + i)\end{align*}$
   ::乘数-6-i)(-6+i)
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3. Multiply: $\begin{align*}\left (\frac{\sqrt{3}}{2} - \frac{1}{2} i \right )^2\end{align*}$
   ::乘数________________________________________________________________________________________________________________________________________________________________
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4. Find the product using polar form: $\begin{align*}(2 + 2i)(\sqrt{3} - i)\end{align*}$
   ::以极表态(2+2i)(%3-i)查找产品
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5. Multiply: $\begin{align*}2(\mbox{cos} \ 40^\circ + i \ \mbox{sin} \ 40^\circ) \bullet 4(\mbox{cos} \ 20^\circ + i \ \mbox{sin} \ 20^\circ)\end{align*}$
   ::乘数: 2 (cos 40i sin 40) 4 (cos 20i sin 20)
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6. Multiply: $\begin{align*}2 \left (\mbox{cos} \ \frac{\pi}{8} + i \ \mbox{sin} \ \frac{\pi}{8} \right ) \bullet 2 \left (\mbox{cos} \ \frac{\pi}{10} + i \ \mbox{sin} \ \frac{\pi}{10} \right )\end{align*}$
   ::乘数: 2(cos 8+i sin 8) 2(cos 10+i sin 10)
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7. Divide: $\begin{align*}2(\mbox{cos} \ 80^\circ + i \ \mbox{sin} \ 80^\circ) \div 6(\mbox{cos} \ 200^\circ + i \ \mbox{sin} \ 200^\circ)\end{align*}$
   ::等号: 2 (cos 80i sin 80) 6 (cos 200i sin 200)
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8. Divide: $\begin{align*}3\ \mbox{cis}(130^\circ) \div 4\ \mbox{cis}(270^\circ)\end{align*}$
   ::除以: 3 Cis( 130) 4 Cis( 270)

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Use De Moivre’s theorem.
::使用德莫伊夫雷的定理。

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9. $\begin{align*}[3(\mbox{cos} \ 80^\circ + i \ \mbox{sin} \ 80^\circ)]^3\end{align*}$
   ::[(cos 80i sin 80]3]
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10. $\begin{align*}\left [\sqrt{2} \left (\mbox{cos}\ \frac{5\pi}{16} + i \ \mbox{sin} \ \frac{5\pi}{16} \right ) \right ]^4\end{align*}$
    ::[2(cos 516+i sin 516)]4]
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11. $\begin{align*}\left (\sqrt{3} - i \right )^6\end{align*}$
    :%3-i)6
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12. Identify the 3 complex cube roots of $\begin{align*}1 + i\end{align*}$
    ::识别 1+i 的 3 个复合立方根
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13. Identify the 4 complex fourth roots of $\begin{align*}-16i\end{align*}$
    ::确定 - 16i 的4个复杂的第四根根
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14. Identify the five complex fifth roots of $\begin{align*}i\end{align*}$
    ::确定i的五个复杂的第五根

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Review (Answers)
::回顾(答复)

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Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键 。