Search
You can find the results of your search below.
Fulltext results:
- Question 4, Exercise 1.3
- 1-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{ali
- Question 2, Exercise 1.4
- n 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}{6
- 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 7, Exercise 1.4
- llowing 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 8, Exercise 1.4
- amplitude 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)\ri
- Question 3, Exercise 1.3
- Question 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 $$a = 1... t{and}\quad c = -48.$$ Then \begin{align} z& = \dfrac{{-6 \pm \sqrt{36-4(1)(-48)}}}{2 \cdot 1} \\ & = \
- 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... n.** \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 4, Exercise 1.1
- al 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+y
- 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 6(x-xvii), Exercise 1.4
- ber 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.// =====Ques... er 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.// =====Q
- 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}\\ =&\dfr
- Question 7, Review Exercise
- egin{align*} &2 z^{2}-11 z+16=0\\ \implies&z^2 - \dfrac{11}{2}z + 8 = 0\\ \implies& z^2 - \dfrac{11}{2}z = -8\\ \implies& z^2 - 2z\dfrac{11}{4}z + \dfrac{121}{16} = -8 + \dfrac{121}{16}\\ \implies&\left(z-\dfrac{11}{4}\right)^2 = -\dfrac{12
- Question 1, Exercise 1.3
- ize 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\righ
- Question 8, Exercise 1.2
- $ by taking $z=x+i y$. **Solution.** Given: $$\dfrac{1}{2} Re(i \bar{z}) = 4.$$ Put $z = x + i y$, then $\bar{z} = x - i y$. We have \begin{align} & \dfrac{1}{2}Re(i(x-iy)) = 4 \\ \implies & \dfrac{1}{2}Re(ix+y)) = 4 \\ \implies & \dfrac{1}{2} y=4 \\ \implies & y=8, \end{align} as required. GOOD ====Q
- Question 2, Review Exercise
- ) ===== Find the value of the following: $\left|\dfrac{(3-2 i)(1+i)}{2-3 i}\right|$ ** Solution. ** \begin{align} &\left|\dfrac{(3-2i)(1+i)}{2-3i}\right|\\ =& \dfrac{|3-2i||1+i|}{|2-3i|}\\ =&\dfrac{ \sqrt{9 + 4}\sqrt{1 + 1}}{\sqrt{4 + 9}}\\ =& \dfrac{\sqrt{13} \cdot \sq