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- Question 4, Exercise 1.3
- -i \end{align} \begin{align} \implies \omega & =\dfrac{1-i}{9+5i}\\ &=\dfrac{1-i}{9+5i}\times\dfrac{9-5i}{9-5i}\\ &=\dfrac{9-5-5i-9i}{81+25}\\ &=\dfrac{4-14i}{106}\\ &=\dfrac{2}{53}-\dfrac{7}{53}i\end{al
- Question 7, Exercise 1.4
- lowing equation in Cartesian form: $\arg (z-1)=-\dfrac{\pi}{4}$ ** Solution. ** Suppose $z=x+iy$, as \begin{align*} &\arg (z-1)=-\dfrac{\pi}{4} \\ \implies & \arg(x+iy-1) = -\dfrac{\pi}{4} \\ \implies & \arg(x-1+iy) = -\dfrac{\pi}{4} \\ \implies & \tan^{-1}\left(\dfrac{y}{x-1}\right)
- Question 9, Exercise 1.2
- . \begin{align} Re(2+4i)^{-1} & = Re(z^{-1}) = \dfrac{Re(z)}{|z|^2} \\ & =\dfrac{2}{2^2+4^2} = \dfrac{2}{20}\\ &= \dfrac{1}{10}. \end{align} \begin{align} Im(2+4i)^{-1} & = Im(z^{-1}) = -\dfrac{Im(z)}{|z
- Question 2, Exercise 1.4
- 2(i)===== Write the complex number $\left(\cos \dfrac{\pi}{6}+i \sin \dfrac{\pi}{6}\right)\left(\cos \dfrac{\pi}{3}+i \sin \dfrac{\pi}{3}\right)$ in rectangular form. ** Solution. ** Let $z_1=\cos \dfrac{\pi}{
- Question 3, Exercise 1.1
- ====Question 3(i)==== Simplify the following $\dfrac{(2+i)(3-2i)}{1+i}$ **Solution.** \begin{align}&\dfrac{(2+i)(3-2i)}{1+i}\\ =&\dfrac{6-2i^2+3i-4i}{1+i}\\ =&\dfrac{8-i}{1+i}\\ =&\dfrac{8-i}{1+i}\times \dfrac{1-i}{1-i}\\ =&\dfrac{8+i^2-8i
- Question 1, Exercise 1.4
- ign} and \begin{align} \alpha & = \tan^{-1}\left|\frac{y}{x}\right| = \tan^{-1}\left|\frac{2\sqrt{3}}{2}\right|\\ & = \tan^{-1}(\sqrt{3}) = \frac{\pi}{3}. \end{align} Since the complex number \(... e argument \( \theta \) is: \[ \theta = \alpha = \frac{\pi}{3}. \] Hence \[ 2 + i 2 \sqrt{3} = 4 \left(
- Question 8, Exercise 1.4
- mplitude is $0.004 \mathrm{~mm}$ and angle is: $\dfrac{\pi}{4}$ ** Solution. ** Here we have $$x_{\max}=0.004, \quad \theta=\dfrac{\pi}{4}.$$ By using the formula \begin{align} x&=x_{\max} e^{i\theta} \\ &=0.004 e^{i\dfrac{\pi}{4}} \\ &=\frac{4}{1000} \left(\cos\left(\dfrac{\pi}{4}\right) +i \sin\left(\dfrac{\pi}{4}\right)\r
- Question 3, Exercise 1.3
- uestion 3(i)==== Solve the quadratic equation: $\dfrac{1}{3} z^{2}+2 z-16=0$. **Solution.** Given \begin{align}&\dfrac{1}{3}z^{2}+2 z-16=0\\ \implies &z^{2} + 6z - 48 =... \end{align} Apply the quadratic formula: $$ z = \dfrac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a},$$ where $$a = 1... {and}\quad c = -48.$$ Then \begin{align} z& = \dfrac{{-6 \pm \sqrt{36-4(1)(-48)}}}{2 \cdot 1} \\ & = \
- Question 6(i-ix), Exercise 1.4
- +i \sin 315^{\circ}\right) \\ =& \sqrt{2} \left(\dfrac{1}{\sqrt{2}}-\dfrac{i}{\sqrt{2}} \right) \\ =& 1-i. \end{align} =====Question 6(ii)===== Write a given ... 10^\circ + i \sin 210^\circ\right) \\ =& 5\left(-\frac{\sqrt{3}}{2} - \frac{1}{2}i\right) \\ =& -\frac{5\sqrt{3}}{2} - \frac{5}{2}i \end{align*} =====Question
- Question 2, Exercise 1.3
- ==== Solve the equation by completing square: $-\dfrac{1}{2} z^{2}-5 z+2=0$. **Solution.** \begin{align} -\dfrac{1}{2} z^{2} - 5z + 2& = 0 \end{align} Multiply th... .** \begin{align} 4z^{2} + 5z &= 14\\ z^{2} + \dfrac{5}{4}z& = \dfrac{14}{4} \\ (z + \dfrac{5}{8})^2 - (\dfrac{5}{8})^2 &=\dfrac{7}{2} \\ (z + \dfrac{5}{8})
- Question 6(x-xvii), Exercise 1.4
- er in the algebraic form: $7 \sqrt{2}\left(\cos \dfrac{5 \pi}{4}+i \sin \dfrac{5 \pi}{4}\right)$ ** Solution. ** //Do yourself as previous parts.// =====Que... r in the algebraic form: $10 \sqrt{2}\left(\cos \dfrac{7 \pi}{4}+i \sin \dfrac{7 \pi}{4}\right)$ ** Solution. ** //Do yourself as previous parts.// =====
- Question 4, Exercise 1.1
- l number $x$ and $y$ in each of the following: $\dfrac{x}{(1+i)}+\dfrac{y}{1-2i}=1$ **Solution.** \begin{align}&\dfrac{x}{(1+i)}+\dfrac{y}{1-2i}=1\\ \implies &\dfrac{x(1-2i)+y(1+i)}{(1+i)(1-2i)}=1\\ \implies &\dfrac{x-i2x+
- Question 10, Exercise 1.2
- +2 i$ and $z_{2}=1-3 i$ verify: $\overline{\left(\frac{z_{1}}{z_{2}}\right)}=\frac{\overline{z_{1}}}{\overline{z_{2}}}$. **Solution.** Given \[z_1 = -3 + 2i, \quad z_2 = 1 - 3i\] Then \begin{align} \frac{z_1}{z_2} &= \frac{-3 + 2i}{1 - 3i}\\ &= \frac{(-3 + 2i)(1 + 3i)}{(1 - 3i)(1 + 3i)} \\ &= \frac{-3 + 2i
- Question 2, Exercise 1.1
- *Solution.** \begin{align}&(3,2)\div(3,-1)\\ =&\dfrac{3+2i}{3-i}\\ =&\dfrac{3+2i}{3-i}\times\dfrac{3+i}{3+i}\\ =&\dfrac{(3+2i)(3+i)}{3^2-i^2}\\ =&\dfrac{9+2i^2+6i+3i}{9+1}\\ =&\dfrac{9-2+9i}{10}\\ =&\df
- Question 1, Exercise 1.3
- ze the polynomial into linear functions: $z^{2}+\dfrac{3}{25}$. **Solution.** \begin{align} & z^2 + \dfrac{3}{25} \\ = & z^2 - \left(\dfrac{\sqrt{3}}{5} i \right)^2 \\ = & \left(z + \dfrac{\sqrt{3}}{5}i\right)\left(z - \dfrac{\sqrt{3}}{5}i\rig