题目
Problem
It is given that z = − 3 + i z=-\sqrt3+\mathrm{i} z = − 3 + i
(a) Express z 2 z^2 z 2 r e i θ r\mathrm{e}^{\mathrm{i}\theta} r e i θ r > 0 r>0 r > 0 − π < θ ≤ π -\pi<\theta\le\pi − π < θ ≤ π
(3)
(b) The complex number ω \omega ω z 2 ω z^2\omega z 2 ω ∣ z 2 ω ∣ = 12 \Big|\dfrac{z^2}{\omega}\Big|=12 ω z 2 = 12
Find the two possible values of ω \omega ω R e i α R\mathrm{e}^{\mathrm{i}\alpha} R e i α R > 0 R>0 R > 0 − π < α ≤ π -\pi<\alpha\le\pi − π < α ≤ π
(3)
题目中文翻译
已知 z = − 3 + i z=-\sqrt3+\mathrm{i} z = − 3 + i
(a) 将 z 2 z^2 z 2 r e i θ r\mathrm{e}^{\mathrm{i}\theta} r e i θ r > 0 r>0 r > 0 − π < θ ≤ π -\pi<\theta\le\pi − π < θ ≤ π
(3)
(b) 复数 ω \omega ω z 2 ω z^2\omega z 2 ω ∣ z 2 ω ∣ = 12 \Big|\dfrac{z^2}{\omega}\Big|=12 ω z 2 = 12
求 ω \omega ω R e i α R\mathrm{e}^{\mathrm{i}\alpha} R e i α R > 0 R>0 R > 0 − π < α ≤ π -\pi<\alpha\le\pi − π < α ≤ π
(3)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程；仅给出计算器答案而没有过程，不予评分。除非题目另有精度要求，非精确数值答案保留 3 位有效数字，以度为单位的角度保留 1 位小数。
解答
(a)
解法一
思路
展开
先求 z z z 2 π 2\pi 2 π
答题过程
展开
For z = − 3 + i z=-\sqrt3+\mathrm{i} z = − 3 + i
∣ z ∣ = 3 + 1 = 2. |z|=\sqrt{3+1}=2. ∣ z ∣ = 3 + 1 = 2. Since z z z
arg z = π − tan − 1 1 3 = 5 π 6 . \arg z=\pi-\tan^{-1}\frac1{\sqrt3}=\frac{5\pi}{6}. arg z = π − tan − 1 3 1 = 6 5 π . Thus
z 2 = 4 e 5 π i / 3 = 4 e − π i / 3 , z^2=4e^{5\pi\mathrm{i}/3}=4e^{-\pi\mathrm{i}/3}, z 2 = 4 e 5 π i /3 = 4 e − π i /3 , where − π / 3 -\pi/3 − π /3
z 2 = 4 e − π i / 3 . \boxed{z^2=4e^{-\pi\mathrm{i}/3}}. z 2 = 4 e − π i /3 .
解法二
思路
展开
也可以直接平方，再从所得实部、虚部判断模和所在象限。
答题过程
展开
Squaring directly,
z 2 = ( − 3 + i ) 2 = 3 − 2 3 i − 1 = 2 − 2 3 i . \begin{align*}
z^2=&\,(-\sqrt3+\mathrm{i})^2\\[4mm]
=&\,3-2\sqrt3\,\mathrm{i}-1\\[4mm]
=&\,2-2\sqrt3\,\mathrm{i}.
\end{align*} z 2 = = = ( − 3 + i ) 2 3 − 2 3 i − 1 2 − 2 3 i . Hence
∣ z 2 ∣ = 2 2 + ( 2 3 ) 2 = 4. |z^2|=\sqrt{2^2+(2\sqrt3)^2}=4. ∣ z 2 ∣ = 2 2 + ( 2 3 ) 2 = 4. The real part is positive and the imaginary part negative, so the principal argument is in the fourth quadrant:
arg ( z 2 ) = − tan − 1 2 3 2 = − π 3 . \arg(z^2)=-\tan^{-1}\frac{2\sqrt3}{2}=-\frac\pi3. arg ( z 2 ) = − tan − 1 2 2 3 = − 3 π . Therefore z 2 = 4 e − π i / 3 \boxed{z^2=4e^{-\pi\mathrm{i}/3}} z 2 = 4 e − π i /3
(b)
解法一
思路
展开
商的模给出 ω \omega ω 0 0 0 π \pi π
答题过程
展开
Let ω = R e i α \omega=Re^{\mathrm{i}\alpha} ω = R e i α
∣ z 2 ω ∣ = 4 R = 12 , R = 1 3 . \bigg|\frac{z^2}{\omega}\bigg|=\frac4R=12,
\qquad R=\frac13. ω z 2 = R 4 = 12 , R = 3 1 . Also
z 2 ω = 4 R e i ( α − π / 3 ) . z^2\omega=4R e^{\mathrm{i}(\alpha-\pi/3)}. z 2 ω = 4 R e i ( α − π /3 ) . Since this product is non-zero and real, α − π / 3 \alpha-\pi/3 α − π /3 π \pi π − π < α ≤ π -\pi<\alpha\le\pi − π < α ≤ π
α = π 3 , α = − 2 π 3 . \alpha=\frac\pi3,\qquad \alpha=-\frac{2\pi}{3}. α = 3 π , α = − 3 2 π . Thus
ω = 1 3 e π i / 3 or ω = 1 3 e − 2 π i / 3 . \boxed{\omega=\frac13e^{\pi\mathrm{i}/3}
\quad\text{or}\quad
\omega=\frac13e^{-2\pi\mathrm{i}/3}}. ω = 3 1 e π i /3 or ω = 3 1 e − 2 π i /3 . 