题目
Problem
The complex number z z z z = 5 a − 2 i 3 + a i z=\dfrac{5a-2\mathrm{i}}{3+a\mathrm{i}} z = 3 + a i 5 a − 2 i a a a arg z = − 1 4 π \arg z=-\frac14\pi arg z = − 4 1 π
(a) Find the value of a a a z z z x + i y x+\mathrm{i}y x + i y x x x y y y
(6)
(b) Express z 3 z^3 z 3 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 r r r θ \theta θ
(3)
题目中文翻译
复数 z z z z = 5 a − 2 i 3 + a i z=\dfrac{5a-2\mathrm{i}}{3+a\mathrm{i}} z = 3 + a i 5 a − 2 i a a a arg z = − 1 4 π \arg z=-\frac14\pi arg z = − 4 1 π
(a) 求 a a a z z z x + i y x+\mathrm{i}y x + i y x x x y y y
(6)
(b) 将 z 3 z^3 z 3 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 r r r θ \theta θ
(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
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先共轭化分母。辐角为 − π / 4 -\pi/4 − π /4 a a a
答题过程
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Multiplying by the conjugate,
z = ( 5 a − 2 i ) ( 3 − a i ) 9 + a 2 = 15 a − 5 a 2 i − 6 i − 2 a 9 + a 2 = 13 a 9 + a 2 − i 5 a 2 + 6 9 + a 2 . \begin{align*}
z
=&\,\frac{(5a-2\mathrm{i})(3-a\mathrm{i})}{9+a^2}\\[4mm]
=&\,\frac{15a-5a^2\mathrm{i}-6\mathrm{i}-2a}{9+a^2}\\[4mm]
=&\,\frac{13a}{9+a^2}
-\mathrm{i}\frac{5a^2+6}{9+a^2}.
\end{align*} z = = = 9 + a 2 ( 5 a − 2 i ) ( 3 − a i ) 9 + a 2 15 a − 5 a 2 i − 6 i − 2 a 9 + a 2 13 a − i 9 + a 2 5 a 2 + 6 . Since arg z = − π / 4 \arg z=-\pi/4 arg z = − π /4
5 a 2 + 6 = 13 a , 5 a 2 − 13 a + 6 = 0. \begin{align*}
5a^2+6=&\,13a,\\[4mm]
5a^2-13a+6=&\,0.
\end{align*} 5 a 2 + 6 = 5 a 2 − 13 a + 6 = 13 a , 0. Factorising,
( 5 a − 3 ) ( a − 2 ) = 0 , (5a-3)(a-2)=0, ( 5 a − 3 ) ( a − 2 ) = 0 , so a = 3 / 5 a=3/5 a = 3/5 a = 2 a=2 a = 2 a a a a = 2 \boxed{a=2} a = 2
Substituting,
z = 26 13 − i 26 13 = 2 − 2 i , z=\frac{26}{13}-\mathrm{i}\frac{26}{13}
=\boxed{2-2\mathrm{i}}, z = 13 26 − i 13 26 = 2 − 2 i , which has the required argument.
解法二
思路
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也可根据辐角先设 z = t − t i z=t-t\mathrm{i} z = t − t i t > 0 t>0 t > 0 t t t
答题过程
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Since arg z = − π / 4 \arg z=-\pi/4 arg z = − π /4 z = t ( 1 − i ) z=t(1-\mathrm{i}) z = t ( 1 − i ) t > 0 t>0 t > 0
5 a − 2 i = t ( 1 − i ) ( 3 + a i ) = t [ ( 3 + a ) + ( a − 3 ) i ] . \begin{align*}
5a-2\mathrm{i}=&\,t(1-\mathrm{i})(3+a\mathrm{i})\\[4mm]
=&\,t\big[(3+a)+(a-3)\mathrm{i}\big].
\end{align*} 5 a − 2 i = = t ( 1 − i ) ( 3 + a i ) t [ ( 3 + a ) + ( a − 3 ) i ] . Comparing parts,
5 a = t ( 3 + a ) , − 2 = t ( a − 3 ) . \begin{align*}
5a=&\,t(3+a),\\[4mm]
-2=&\,t(a-3).
\end{align*} 5 a = − 2 = t ( 3 + a ) , t ( a − 3 ) . The second gives t ( 3 − a ) = 2 t(3-a)=2 t ( 3 − a ) = 2 t t t
5 a ( 3 − a ) = 2 ( 3 + a ) . 5a(3-a)=2(3+a). 5 a ( 3 − a ) = 2 ( 3 + a ) . Thus
5 a 2 − 13 a + 6 = 0 , ( 5 a − 3 ) ( a − 2 ) = 0. \begin{align*}
5a^2-13a+6=&\,0,\\[4mm]
(5a-3)(a-2)=&\,0.
\end{align*} 5 a 2 − 13 a + 6 = ( 5 a − 3 ) ( a − 2 ) = 0 , 0. The integer condition selects a = 2 \boxed{a=2} a = 2 t = 2 / ( 3 − 2 ) = 2 t=2/(3-2)=2 t = 2/ ( 3 − 2 ) = 2 z = 2 − 2 i \boxed{z=2-2\mathrm{i}} z = 2 − 2 i
(b)
解法一
思路
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先求 z z z
答题过程
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From part (a), z = 2 − 2 i z=2-2\mathrm{i} z = 2 − 2 i
∣ z ∣ = 2 2 + ( − 2 ) 2 = 2 2 , arg z = − π 4 . |z|=\sqrt{2^2+(-2)^2}=2\sqrt2,
\qquad \arg z=-\frac\pi4. ∣ z ∣ = 2 2 + ( − 2 ) 2 = 2 2 , arg z = − 4 π . Hence
∣ z 3 ∣ = ( 2 2 ) 3 = 16 2 , arg ( z 3 ) = 3 ( − π 4 ) = − 3 π 4 . \begin{align*}
|z^3|=&\,(2\sqrt2)^3=16\sqrt2,\\[4mm]
\arg(z^3)=&\,3\bigg(-\frac\pi4\bigg)=-\frac{3\pi}{4}.
\end{align*} ∣ z 3 ∣ = arg ( z 3 ) = ( 2 2 ) 3 = 16 2 , 3 ( − 4 π ) = − 4 3 π . This argument is in ( − π , π ] (-\pi,\pi] ( − π , π ]
z 3 = 16 2 e − 3 π i / 4 . \boxed{z^3=16\sqrt2\,e^{-3\pi\mathrm{i}/4}}. z 3 = 16 2 e − 3 π i /4 . Thus r = 16 2 r=\boxed{16\sqrt2} r = 16 2 θ = − 3 π / 4 \theta=\boxed{-3\pi/4} θ = − 3 π /4
