题目
Problem
The complex number z 1 z_1 z 1
z 1 = r ( cos 7 π 6 + i sin 7 π 6 ) \begin{align*}
z_1
=&\,r\big(\cos \tfrac{7\pi}{6}\\
&\,\hspace{2pt}+\mathrm{i}\sin \tfrac{7\pi}{6}\big)
\end{align*} z 1 = r ( cos 6 7 π + i sin 6 7 π ) where r r r
The complex number z 2 z_2 z 2
Given that ∣ z 1 z 2 ∣ = 15 |z_1 z_2| = 15 ∣ z 1 z 2 ∣ = 15
(a) state the value of r r r
(1)
Given further that z 1 + z 2 z_1 + z_2 z 1 + z 2
(b) determine the possible complex numbers z 2 z_2 z 2 a + b i a + b\mathrm{i} a + b i a a a b b b
(5)
(c) Show z 1 z_1 z 1 z 2 z_2 z 2
(2)
题目中文翻译
复数 z 1 z_1 z 1
z 1 = r ( cos 7 π 6 + i sin 7 π 6 ) \begin{align*}
z_1
=&\,r\big(\cos \tfrac{7\pi}{6}\\
&\,\hspace{2pt}+\mathrm{i}\sin \tfrac{7\pi}{6}\big)
\end{align*} z 1 = r ( cos 6 7 π + i sin 6 7 π ) 其中 r r r
复数 z 2 z_2 z 2
已知 ∣ z 1 z 2 ∣ = 15 |z_1 z_2| = 15 ∣ z 1 z 2 ∣ = 15
(a) 写出 r r r
(1)
进一步已知 z 1 + z 2 z_1 + z_2 z 1 + z 2
(b) 求可能的复数 z 2 z_2 z 2 a + b i a + b\mathrm{i} a + b i a a a b b b
(5)
(c) 在同一幅 Argand 图上画出 z 1 z_1 z 1 z 2 z_2 z 2
(2)
解答
(a)
解法一
思路
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复数乘积的模等于两复数模的乘积。这里 ∣ z 1 ∣ = r |z_1|=r ∣ z 1 ∣ = r ∣ z 2 ∣ = 5 |z_2|=5 ∣ z 2 ∣ = 5
答题过程
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Using ∣ z 1 z 2 ∣ = ∣ z 1 ∣ ∣ z 2 ∣ |z_1z_2|=|z_1||z_2| ∣ z 1 z 2 ∣ = ∣ z 1 ∣∣ z 2 ∣
5 r = 15. 5r=15. 5 r = 15. Since r r r
r = 3 . \boxed{r=3}. r = 3 .
(b)
解法一
思路
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先把 z 1 z_1 z 1 z 2 = a + b i z_2=a+b\mathrm{i} z 2 = a + b i z 1 + z 2 z_1+z_2 z 1 + z 2 b b b ∣ z 2 ∣ = 5 |z_2|=5 ∣ z 2 ∣ = 5
答题过程
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From part (a), r = 3 r=3 r = 3
cos 7 π 6 = − 3 2 \cos\frac{7\pi}6=-\frac{\sqrt3}2 cos 6 7 π = − 2 3 and
sin 7 π 6 = − 1 2 . \sin\frac{7\pi}6=-\frac12. sin 6 7 π = − 2 1 . Therefore,
z 1 = 3 ( cos 7 π 6 + i sin 7 π 6 ) = 3 ( − 3 2 − 1 2 i ) = − 3 3 2 − 3 2 i . \begin{align*}
z_1
=&\,3\bigg(\cos\frac{7\pi}{6}
+\mathrm{i}\sin\frac{7\pi}{6}\bigg)\\
=&\,3\bigg(-\frac{\sqrt3}2-\frac12\mathrm{i}\bigg)\\
=&\,-\frac{3\sqrt3}{2}-\frac32\mathrm{i}.
\end{align*} z 1 = = = 3 ( cos 6 7 π + i sin 6 7 π ) 3 ( − 2 3 − 2 1 i ) − 2 3 3 − 2 3 i . Let z 2 = a + b i z_2=a+b\mathrm{i} z 2 = a + b i z 1 + z 2 z_1+z_2 z 1 + z 2
− 3 2 + b = 0 b = 3 2 . \begin{align*}
-\frac32+b=&\,0\\
b=&\,\frac32.
\end{align*} − 2 3 + b = b = 0 2 3 . Also, ∣ z 2 ∣ = 5 |z_2|=5 ∣ z 2 ∣ = 5
a 2 + b 2 = 25 a 2 + 9 4 = 25 a 2 = 91 4 . \begin{align*}
a^2+b^2=&\,25\\
a^2+\frac94=&\,25\\
a^2=&\,\frac{91}{4}.
\end{align*} a 2 + b 2 = a 2 + 4 9 = a 2 = 25 25 4 91 . Thus a = ± 91 2 a=\pm\frac{\sqrt{91}}{2} a = ± 2 91
z 2 = 91 2 + 3 2 i \boxed{
z_2=\frac{\sqrt{91}}{2}+\frac32\mathrm{i}
} z 2 = 2 91 + 2 3 i or
z 2 = − 91 2 + 3 2 i . \boxed{
z_2=-\frac{\sqrt{91}}{2}+\frac32\mathrm{i}
}. z 2 = − 2 91 + 2 3 i .
解法二
思路
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也可以使用三角形式。把 z 2 z_2 z 2 sin θ \sin\theta sin θ sin 2 θ + cos 2 θ = 1 \sin^2\theta+\cos^2\theta=1 sin 2 θ + cos 2 θ = 1
答题过程
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Write
z 2 = 5 ( cos θ + i sin θ ) . z_2=5(\cos\theta+\mathrm{i}\sin\theta). z 2 = 5 ( cos θ + i sin θ ) . Since the imaginary part of z 1 + z 2 z_1+z_2 z 1 + z 2
3 sin 7 π 6 + 5 sin θ = 0. 3\sin\frac{7\pi}{6}+5\sin\theta=0. 3 sin 6 7 π + 5 sin θ = 0. Therefore,
− 3 2 + 5 sin θ = 0 5 sin θ = 3 2 sin θ = 3 10 . \begin{align*}
-\frac32+5\sin\theta=&\,0\\
5\sin\theta=&\,\frac32\\
\sin\theta=&\,\frac3{10}.
\end{align*} − 2 3 + 5 sin θ = 5 sin θ = sin θ = 0 2 3 10 3 . Hence
cos θ = ± 1 − ( 3 10 ) 2 = ± 91 10 . \begin{align*}
&\,\cos\theta\\
=&\,\pm\sqrt{1-\bigg(\frac3{10}\bigg)^2}\\
=&\,\pm\frac{\sqrt{91}}{10}.
\end{align*} = = cos θ ± 1 − ( 10 3 ) 2 ± 10 91 . It follows that
z 2 = 5 ( ± 91 10 + 3 10 i ) = ± 91 2 + 3 2 i . \begin{align*}
z_2
=&\,5\bigg(
\pm\frac{\sqrt{91}}{10}
+\frac3{10}\mathrm{i}\bigg)\\
=&\,\boxed{
\pm\frac{\sqrt{91}}{2}+\frac32\mathrm{i}
}.
\end{align*} z 2 = = 5 ( ± 10 91 + 10 3 i ) ± 2 91 + 2 3 i .
(c)
解法一
思路
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在同一 Argand 图上标出三个复数：z 1 z_1 z 1 z 2 z_2 z 2 ∣ z 1 ∣ < ∣ z 2 ∣ |z_1|<|z_2| ∣ z 1 ∣ < ∣ z 2 ∣ z 1 z_1 z 1
答题过程
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The three points to be shown are
z 1 = − 3 3 2 − 3 2 i , z_1=-\frac{3\sqrt3}{2}-\frac32\mathrm{i}, z 1 = − 2 3 3 − 2 3 i , and
z 2 = 91 2 + 3 2 i , z_2=\frac{\sqrt{91}}{2}+\frac32\mathrm{i}, z 2 = 2 91 + 2 3 i , and
z 2 = − 91 2 + 3 2 i . z_2=-\frac{\sqrt{91}}{2}+\frac32\mathrm{i}. z 2 = − 2 91 + 2 3 i . Thus z 1 z_1 z 1 z 2 z_2 z 2
