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- Question 7, Exercise 1.2
- i)===== Separate into real and imaginary parts $\dfrac{2+3i}{5-2i}$. ====Solution==== \begin{align}&\dfrac{2+3i}{5-2i} \\ =&\dfrac{2+3i}{5-2i}\times \dfrac{5+2i}{5+2i} \quad \text{by rationalizing} \\ =&\dfrac{10-6+15i+4i}{25+4}\\ =&\
- Question 8, Exercise 1.1
- Pakistan. =====Question 8(i)===== Express the $\dfrac{1-2i}{2+i}+\dfrac{4-i}{3+2i}$ in the standard form of $a+ib.$ ====Solution==== \begin{align}&\dfrac{1-2i}{2+i}+\dfrac{4-i}{3+2i}\\ &=\dfrac{\left( 3+2i \right)\left( 1-2i \right)+\left( 2+i \right)\left(
- Question 6, Exercise 1.1
- stion 6(i)===== Perform the indicated division $\dfrac{4+i}{3+5i}$ and write the answer in the form $a+ib$. ====Solution==== \begin{align}\dfrac{4+i}{3+5i}&=\dfrac{4+i}{3+5i}\times \dfrac{3-5i}{3-5i}\\ &=\dfrac{\left( 12+5 \right)+\left( 3-20 \right)i}{9-25{{i}^{2}}}
- Question 1, Review Exercise 1
- se the correct option. <panel> i. ${{\left( \dfrac{2i}{1+i} \right)}^{2}}$ * (a) $i$ * (b... apsed="true">(B): $2i$</collapse> ii. Divide $\dfrac{5+2i}{4-3i}$ * (a) $-\dfrac{7}{25}+\dfrac{26}{25}i$ * (b) $\dfrac{5}{4}-\dfrac{2}{3}i$ * %%(c)%% $\dfrac{14}{25}+\dfrac{2
- Question 6, Exercise 1.3
- 1 )=0$$ By using quadratic formula, we have $$z=\dfrac{-1\pm \sqrt{1-4}}{2}$$ $$z=\dfrac{-1\pm \sqrt{3}i}{2}$$ $$(z^2-z+1 )=0$$ By using quadratic formula, we have $$z=\dfrac{1\pm \sqrt{1-4}}{2}$$ $$z=\dfrac{1\pm \sqrt{3}i}{2}$$ The value of $z$ from both equations, we have $$z
- Question 6, 7 & 8, Review Exercise 1
- istan. =====Question 6===== Find conjugate of $\dfrac{1}{3+4i}$.\\ ====Solution==== Suppose $$z=\dfrac{1}{3+4i}.$$ Then \begin{align}z&=\dfrac{1}{3+4i}\times \dfrac{3-4i}{3-4i}\\ &=\dfrac{3-4i}{9+16}\\ &=\dfrac{3-4i}{25}\\ &=\dfrac{3}{25}-\dfrac{
- Question 6, Exercise 1.2
- numbers ${{z}_{1}}$and ${{z}_{2}}$that $\left| \dfrac{{{z}_{1}}}{{{z}_{2}}} \right|=\dfrac{|{{z}_{1}}|}{|{{z}_{2}}|}$, where ${{z}_{2}}\ne 0$ ====Solution==... |=\sqrt{a^2+b^2}$. We take \begin{align}\left| \dfrac{1}{z} \right|&=\left| \dfrac{1}{a+bi} \right|\\ &=\left| \dfrac{1}{a+bi}\cdot \dfrac{a-bi}{a-bi} \right
- Question 4 & 5, Review Exercise 1
- f ${{z}_{1}}=2-i$, ${{z}_{2}}=1+i,$ find $\left|\dfrac{{{z}_{1}}+{{z}_{2}}+1}{{{z}_{1}}-{{z}_{2}}+1}\rig... _1=2-i$ and $z_2=1+i$, so we have \begin{align} \dfrac{{{z}_{1}}+{{z}_{2}}+1}{{{z}_{1}}-{{z}_{2}}+1}&=\dfrac{\left( 2-i \right)+\left( 1+i \right)+1}{\left( 2-i \right)-\left( 1+i \right)+1}\\ &=\dfrac{2-i+1+i+1}{2-i-1-i+1}\\ &=\dfrac{4}{2-2i}\\ &=\df
- Question 9 & 10, Exercise 1.1
- . =====Question 9===== Find the conjugate of $\dfrac{\left( 3-2i \right)\left( 2+3i \right)}{\left( 1+... \right)}$. ====Solution==== Let \begin{align}z&=\dfrac{\left( 3-2i \right)\left( 2+3i \right)}{\left( 1+2i \right)\left( 2-i \right)}\\ &=\dfrac{6+6+9i-4i}{2+2+4i-i}\\ &=\dfrac{12+5i}{4+3i}\end{align} Now \begin{align}\bar{z}&=\dfrac{12-5i}{4-3i}\\
- Question 5, Exercise 1.2
- {{z}_{2}}=2a-3bi$. Verify that $\overline{\left(\dfrac{{{z}_{1}}}{{{z}_{2}}}\right)}=\dfrac{\overline{{{z}_{1}}}}{\overline{{{z}_{2}}}}$. ====Solution==== Gi... and $\overline{z_2}=2a+3bi$. Take \begin{align}\dfrac{z_1}{z_2}&=\dfrac{-a-3bi}{2a-3bi}\\ &=\dfrac{-a-3bi}{2a-3bi}\times \dfrac{2a+3bi}{2a+3bi} \quad (\text{
- Question 3 & 4, Exercise 1.2
- $5+2i$ is $-5-2i$. Now \begin{align}z^{-1} &= \dfrac{a}{{{a}^{2}}+{{b}^{2}}}-\dfrac{b}{{{a}^{2}}+{{b}^{2}}}i\\ &=\dfrac{5}{5^2+2^2}-\dfrac{2}{5^2+2^2}i\\ &=\dfrac{5}{29}-\dfrac{2}{29}i \end{align} Thus multiplicative invers
- Question 2 & 3, Exercise 1.1
- ===Question 3(ii)===== Add the complex numbers $\dfrac{1}{2}-\dfrac{2}{3}i,\dfrac{1}{4}-\dfrac{1}{3}i$. ====Solution==== \begin{align}&\left( \dfrac{1}{2}-\dfrac{2}{3}i \right)+\left( \
- Question 7, Exercise 1.1
- 1}}=1+2i$ and ${{z}_{2}}=2+3i$, evaluate $\left|\dfrac{z_1}{z_2}\right|$. ====Solution==== We know that $z_1=1+2i$ and $z_2=2+3i$, then \begin{align} \dfrac{z_1}{z_2}&=\dfrac{1+2i}{2+3i}\\ &=\dfrac{1+2i}{2+3i}\times \dfrac{2-3i}{2-3i}\\ &=\dfrac{\left( 2+6 \right)+\left( 4-3 \r
- Question 11, Exercise 1.1
- }}=2-i$, ${{z}_{2}}=-2+i$. Find ${\rm Re}\left( \dfrac{{{z}_{1}}{{z}_{2}}}{\overline{{{z}_{1}}}} \right)... \ &=-3+4i \end{align} Now we take \begin{align} \dfrac{z_1 z_2}{\overline{{z_1}}}&=\dfrac{-3+4i}{2+i}\\ &=\dfrac{-3+4i}{2+i}\times \dfrac{2-i}{2-i}\\ &=\dfrac{-6+4+8i+3i}{4+1}\\ &=\dfrac{-2+11i
- Question 5, Exercise 1.3
- b=1$ and $c=3$.\\ Thus we have \begin{align}z&=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\ &=\dfrac{-1\pm \sqrt{{{\left( 1 \right)}^{2}}-4\left( 1 \right)\left( 3 \right)}}{2\left( 1 \right)}\\ &=\dfrac{-1\pm \sqrt{1-12}}{2}\\ &=\dfrac{-1\pm \sqrt{11}}{2}i\end{align} Thus the solutions of the given equati