Example 4
Calculate
$\frac{2 + 3i}{3 + i}$.
Solution
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Exercise 1.3
Question 1
Let $z_1=-2+3i,z_2=5+2i.$ Compute:(a) $z_1^2- 2z_1+ 1$
Short Answer$ -18i $
(b) $3z_2^2+2z_2-1$
Short Answer$ 72+64i$
(c) $z_1\overline{z_2} +z_2\overline{z_1}$
Short Answer$-8 $
(d) $\frac 1{z_1}$
Short Answer$ -\frac{2}{13} - \frac{3}{13}i $
(e) $\frac 1{z_2}$
Short Answer$ \frac{5}{29} - \frac{2}{29}i $
(f) $\frac 1{z_1z_2}$
Short Answer$ -\frac{16}{377} - \frac{11}{377}i $
(g) ${\frac {z_1}{z_2}}$
Short Answer$ -\frac{4}{29} + \frac{19}{29}i $
(h) $\frac {\overline {z_1}}{\overline {z_2}}$
Short Answer$ -\frac{4}{29} - \frac{19}{29}i $
(i) $\frac {z_2}{z_1}$
Short Answer$-\frac{4}{13} - \frac{19}{13}i $
(j) $\overline {\left ( \frac {z_2}{z_1}\right ) }$
Short Answer$-\frac{4}{13} + \frac{19}{13}i $
(k) $\frac {\overline {z_1}z_2}{z_1\overline {z_2}}$
Short Answer$ -\frac{345}{377} + \frac{152}{377}i $
(l) $\frac {z_2}{\overline {z_1}}+ \frac {z_1}{\overline {z_2}}$
Short Answer$ -\frac{672}{377} + \frac{462}{377}i $
Question 2
Let $z_1= 3- 2i$, $z_{2}= - 1+ 4i.$ Show that(a) $\overline {\left ( z_1+ z_2\right ) }= \overline {z_1}+ \overline {z_2}$
Short Answer$ 2-2i $
(b) $\overline {z_1z_2}= \overline {z_1}$ $\overline {z_2}$
Short Answer$ 5-14i $
(c) $\overline {\left ( \frac {z_1}{z_2}\right ) }= \frac {\overline {z_1}}{\overline {z_2}}$
Short Answer$ -\dfrac{11}{17} + \dfrac{10}{17}i $
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Example 5
Find the trigonometric form with $-\pi \lt \theta \leq \pi$.
(a) $\quad z = 1 + \sqrt{3}i$
(b) $\quad z = -1 + i$
(c) $\quad z = -\sqrt{3} - i$
(d) $\quad z = -1$
Solution
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Example 6
Given that $z_1 = 1 + \sqrt{3}i$, $z_2 = -1 + i$, find $z_1 z_2$ by using trigonometric forms. Check your answer by direct multiplication.
Solution
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Example 7
Given that $z = -\sqrt{3} - i$, using trigonometric form of $z$, find $z^{-1}$. Check your answer by showing that $zz^{-1} = 1$.
Solution
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Example 8
Given that $z_1 = 1 + \sqrt{3}i$, $z_2 = -1 + i$, find $\dfrac{z_1}{z_2}$ by using trigonometric forms. Check your answer by direct calculation.
Solution
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Example 9
Given that $z = 1 + \sqrt{3}i$, find
(a) $z^{10}$
(b) $z^{-10}$.
Solution
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Exercise 1.4
Question 1.
Find the trigonometric form with $-\pi\lt\theta \leq \pi$.(a) $z=1-\sqrt{3} i$
Short Answer$ 2\left(\cos\left(-\frac{\pi}{3}\right) + i \sin\left(-\frac{\pi}{3}\right)\right) $
(b) $z=-\sqrt{2}+\sqrt{2} i$
Short Answer$ 2\left(\cos\frac{3\pi}{4} + i \sin\frac{3\pi}{4}\right) $
(c) $z=-2-2 i$
Short Answer$ 2\sqrt{2}\left(\cos\left(-\frac{3\pi}{4}\right) + i \sin\left(-\frac{3\pi}{4}\right)\right) $
(d) $z=\sqrt{3}-i$
Short Answer$ 2\left(\cos\left(-\frac{\pi}{6}\right) + i \sin\left(-\frac{\pi}{6}\right)\right) $
(e) $z=i$
Short Answer$ \cos\frac{\pi}{2} + i \sin\frac{\pi}{2} $
(f) $z=-3 i$
Short Answer$3\left(\cos\left(-\frac{\pi}{2}\right) + i \sin\left(-\frac{\pi}{2}\right)\right) $
Question 2.
Given that $z_1=2-2 \sqrt{3} i, z_2=-1-i$, find the following complex numbers by using trigonometric forms. Check your answer by direct calculation.(a) $z_1 z_2$
Short Answer$ -2\sqrt{3} - 2 + i(2\sqrt{3} - 2) $
(b) $z_1^{-1}$
Short Answer$ -2\sqrt{3} - 2 + i(2\sqrt{3} - 2) $
(c) $z_2^{-1}$
Short Answer$ -\frac{1}{2} + \frac{1}{2}i $
(d) $\frac{z_1}{z_2}$
Short Answer$ (\sqrt{3} - 1) + i(\sqrt{3} + 1) $
(e) $\frac{z_2}{z_1}$
Short Answer$ \frac{\sqrt{3} - 1}{8} - i\frac{\sqrt{3} + 1}{8} $
Question 3.
Given that $z=-2 \sqrt{3}-2 i$, find(a) $z^5$
Short Answer$512\sqrt{3} - 512i $
(b) $z^{-5}$.
Short Answer$ \frac{\sqrt{3}}{2048} + \frac{1}{2048}i $
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Example 10
Find the cube roots of $z = -2 - 2i$.
Solution
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Example 11
Solve $z^6 = 1$.
Solution
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Exercise 1.5
Question 1.
Find the square roots of the following complex numbers.(a) $1+\sqrt{3} i$
Short Answer$ \pm\left( \frac{\sqrt{6}}{2} + \frac{\sqrt{2}}{2}i \right) $
(b) $i$
Short Answer$ \pm\left( \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i \right) $
(c) $-\sqrt{3}+i$
Short Answer$ \pm\left( \frac{\sqrt{3}-1}{2} + \frac{\sqrt{3}+1}{2}i \right) $
(d) $-1-\sqrt{3} i$
Short Answer$ \pm\left( \frac{\sqrt{2}}{2} - \frac{\sqrt{6}}{2}i \right) $
(e) $-i$
Short Answer$ \pm\left( \frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}i \right) $
(f) $\sqrt{3}-i$
Short Answer$ \pm\left( \frac{\sqrt{3}+1}{2} - \frac{\sqrt{3}-1}{2}i \right) $
Question 2.
Find the cube roots of the following complex numbers.(a) $1+i$
Short Answer$ \begin{aligned}
w_0 &= 2^{1/6}\left( \frac{\sqrt{6}+\sqrt{2}}{4} + i\frac{\sqrt{6}-\sqrt{2}}{4} \right) \\
w_1 &= 2^{1/6}\left( -\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2} \right) \\
w_2 &= 2^{1/6}\left( -\frac{\sqrt{6}-\sqrt{2}}{4} - i\frac{\sqrt{6}+\sqrt{2}}{4} \right)
\end{aligned} $
(b) $i$
Short Answer$ \boxed{\frac{\sqrt{3}}{2} + \frac{1}{2}i,\quad -\frac{\sqrt{3}}{2} + \frac{1}{2}i,\quad -i} $
(c) $-1+i$
Short Answer$ \boxed{\begin{aligned}
w_0 &= 2^{1/6}\left( \frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2} \right) \\
w_1 &= 2^{1/6}\left( -\frac{\sqrt{6}+\sqrt{2}}{4} + i\frac{\sqrt{6}-\sqrt{2}}{4} \right) \\
w_2 &= 2^{1/6}\left( \frac{\sqrt{6}-\sqrt{2}}{4} - i\frac{\sqrt{6}+\sqrt{2}}{4} \right)
\end{aligned}} $
(d) $-1-i$
Short Answer$ \boxed{\begin{aligned}
w_0 &= 2^{1/6}\left( \frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2} \right) \\
w_1 &= 2^{1/6}\left( \frac{\sqrt{6}-\sqrt{2}}{4} + i\frac{\sqrt{6}+\sqrt{2}}{4} \right) \\
w_2 &= 2^{1/6}\left( -\frac{\sqrt{6}+\sqrt{2}}{4} - i\frac{\sqrt{6}-\sqrt{2}}{4} \right)
\end{aligned}} $
(e) $-i$
Short Answer$ \boxed{\frac{\sqrt{3}}{2} - \frac{1}{2}i,\quad i,\quad -\frac{\sqrt{3}}{2} - \frac{1}{2}i} $
(f) $1-i$
Short Answer$ \boxed{\begin{aligned}
w_0 &= 2^{1/6}\left( \frac{\sqrt{6}+\sqrt{2}}{4} - i\frac{\sqrt{6}-\sqrt{2}}{4} \right) \\
w_1 &= 2^{1/6}\left( -\frac{\sqrt{6}-\sqrt{2}}{4} + i\frac{\sqrt{6}+\sqrt{2}}{4} \right) \\
w_2 &= 2^{1/6}\left( -\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2} \right)
\end{aligned}} $
Question 3.
Solve the following equations.(a) $z^4=-i$
Short Answer$ \boxed{z = \pm \frac{\sqrt{2+\sqrt{2}}}{2} \mp i\frac{\sqrt{2-\sqrt{2}}}{2},\; \pm \frac{\sqrt{2-\sqrt{2}}}{2} \pm i\frac{\sqrt{2+\sqrt{2}}}{2}} $
(b) $z^4=-1$
Short Answer$ \boxed{z = \pm \frac{\sqrt{2}}{2} \pm i\frac{\sqrt{2}}{2}}
$
(c) $z^4=-8-8 \sqrt{3} i$
Short Answer$ \boxed{z = \sqrt{3} - i,\; 1 + i\sqrt{3},\; -\sqrt{3} + i,\; -1 - i\sqrt{3}} $
(d) $z^6=-1$
Short Answer$ \boxed{z = \pm \frac{\sqrt{3}}{2} \pm \frac{1}{2}i,\; \pm i} $
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