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- Question 1, Exercise 1.3
- z+3w&=11-5i …(ii)\end{align} Multiply $2$ by (i), we get\\ \begin{align}2z-8w&=6i …(iii)\end{align} Subtract (iii) from (ii), we get\\ \[\begin{array}{cccc} 2z&-8w&=6i \\ \mat... \ 2z+3w&=2 …(ii)\end{align} Multiply $2$ by (i), we get\\ \begin{align}2z+2w&=6i …(iii)\end{align} Subtract (iii) from (ii), we get\\ $$\begin{array}{ccc} 2z & +2w & =6i\\ \ma
- Question 1, Exercise 1.2
- == Given ${{z}_{1}}=2+i$, ${{z}_{2}}=1-i$. First, we prove commutative property under addition, that is, $$z_1+z_2=z_2+z_1.$$ We take \begin{align}z_1+z_2&=(2+i)+(1-i)\\ &=3 \ld... )\\ &=3 \ldots (ii)\end{align} From (i) and (ii), we get required result. Now, we prove commutative property under multiplication, that is, $$z_1 z_2=z_2
- Question 2, Exercise 1.2
- i$, ${{z}_{2}}=3-2i$ and ${{z}_{3}}=2-2i$. First, we prove associative property under addition, that i... \ &=4-3i \ldots (2) \end{align} From (1) and (2), we have the required result. Now, we prove associative property under multiplication, that is, $$z_1 (z... +10)+(2+10)i\\ &=8+12i \ldots (3)\end{align} Now, we take \begin{align} z_1 z_2 &=(-1+i)\cdot (3-2i)\\
- Question 8, Exercise 1.2
- }^{2}}. \ldots (2) \end{align} Using (1) and (2), we get $$z\overline{z}={{\left[\operatorname{Re}\le... ose $z=a+bi$ ... (1) Then $\overline{z}=a-bi$. We have given \begin{align}&z=\overline{z} \\ \impli... i=0 \\ \implies &b=0 \end{align} Using it in (1), we get $z=a+0i=a$, that is, $z$ is real. =====Q... 2a=0\\ \implies &a=0 \end{align} Using it in (1), we get $$z=0+bi=bi,$$ that is, $z$ is pure imaginary
- Question 5, Exercise 1.3
- {2}}+z+3=0$\\ According to the quadratic formula, we have\\ $a=1,\,\,\,b=1$ and $c=3$\\ Quadratic for... {2}}-z-1=0$\\ According to the quadratic formula, we have\\ $a=1,\quad b=-1$ and $c=-1$\\ Quadratic f... 2}}-2z+i=0$\\ According to the quadratic formula, we have\\ $a=1,\,\,\,b=-2$ and $c=i$\\ Quadratic fo... }^{2}}+4=0$\\ According to the quadratic formula, we have\\ $a=1,\,\,\,b=0$ and $c=4$\\ Quadratic form
- Question 7, Exercise 1.1
- valuate $|{{z}_{1}}+{{z}_{2}}|$. ====Solution==== We know that $z_1=1+2i$ and $z_2=2+3i$, then \begin{... evaluate $|{{z}_{1}}{{z}_{2}}|$. ====Solution==== We know that $z_1=1+2i$ and $z_2=2+3i$, then \begin{... $\left|\dfrac{z_1}{z_2}\right|$. ====Solution==== We know that $z_1=1+2i$ and $z_2=2+3i$, then \begin{
- Question 11, Exercise 1.1
- )(-2+i)\\ &=-4+1+2i+2i\\ &=-3+4i \end{align} Now we take \begin{align} \dfrac{z_1 z_2}{\overline{{z_1... &=\dfrac{-2}{5}+\dfrac{11i}{5} \end{align} Hence, we have $${\rm Re}\left( \dfrac{{{z}_{1}}{{z}_{2}}}{... frac{1}{z_1\overline{z_1}}=\dfrac{1}{5}.$$ Hence, we have $${\rm Im}\left(\dfrac{1}{z_1\overline{z_1}}
- Question 5, Exercise 1.2
- i\\ &=3-i \ldots (2)\end{align} From (1) and (2), we have $$\overline{z_1+z_2}=\overline{z_1}+\overli... ^2\\ &=13 \ldots (2)\end{align} From (1) and (2), we have $$\overline{z_1 z_2}=\overline{z_1}\overline... b}^{2}}} \ldots (2) \end{align} From (1) and (2), we have $$\overline{\left(\dfrac{z_1}{z_2}\right)}=\
- Question 6, Exercise 1.2
- 2}}|=R.H.S. \end{align} **Alternative Method**\\ We know $|z|^2=z\bar{z}$, so we have \begin{align}|{{z}_{1}}{{z}_{2}}{{|}^{2}}&={{z}_{1}}{{z}_{2}}\overl... === Suppose $z=a+bi$, then $|z|=\sqrt{a^2+b^2}$. We take \begin{align}\left| \dfrac{1}{z} \right|&=\
- Question 6, Exercise 1.3
- \right)=0$\\ According to the quadratic formula, we have\\ $a=1,\,\,\,b=-2$ and $c=4$ \\ Quadratic fo... 2}}-3z+3=0$\\ According to the quadratic formula, we have\\ $a=1,\,\,\,b=-3$ and $c=3$ \\ Quadratic fo... =0\end{align} According to the quadratic formula, we have\\ $a=1,\quad\quad b=1$ and $c=1$ \\ Quadrati
- Question 2, Exercise 1.3
- ${{z}^{3}}+6z+20$.\\ By using synthetic division, we have $$\begin{array}{c|cccc} -2 & 1 & 0 & 6 & 20 ... }-2{{z}^{2}}+z-2$.\\ By using synthetic division, we have $$\begin{array}{c|cccc} 2 & 1 & -2 & 1 & -2
- Question 3 & 4, Exercise 1.3
- ied the equation. =====Question 4===== Determine weather $1+2i$ is a solution of ${{z}^{2}}-2z+5=0$\\... 2}}-2z+5=0$\\ According to the quadratic formula, we have\\ $a=1,\quad b=-2$ and $c=5$\\ Quadratic fo
- Question 6, 7 & 8, Review Exercise 1
- =0\end{align} According to the quadratic formula, we have\\ $$a=1,\quad b=-2$$ and $$c=2$$\\ Quadrati