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- Question 3 & 4, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- satisfied the equation ${{z}^{2}}+2z+2=0$\\ ====Solution==== Given: $$z^2+2z_1+2=0\quad \ldots (i)$$ Put ... ====Question 4===== Determine weather $1+2i$ is a solution of ${{z}^{2}}-2z+5=0$\\ ====Solution==== ${{z}^{2}}-2z+5=0$\\ According to the quadratic formula, we have\
- Question 2, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- actors. $$P\left( z \right)={{z}^{3}}+6z+20$$ ====Solution==== Given: $$p\left( z \right)={{z}^{3}}+6z+20$... $P(z)$ into linear factors. $$P(z)=3z^2+7.$$ ====Solution==== \begin{align} P(z)&=3z^2+7\\ &=\left(\sqrt{3}... r factors. $$P\left( z \right)={{z}^{2}}+4$$ ====Solution==== \begin{align}P(z)&={{z}^{2}}+4\\ &={{\left( ... r factors. $$P(z)={{z}^{3}}-2{{z}^{2}}+z-2.$$ ====Solution==== Given: $$P\left( z \right)={{z}^{3}}-2{{z}^{
- Question 1, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- in{align}&z-4w=3i\\ &2z+3w=11-5i\end{align} ====Solution==== Given that \begin{align}z-4w&=3i …(i)\\ 2z... t. \begin{align}&z+w=3i\\ &2z+3w=2\end{align} ====Solution==== Given that \begin{align}z+w&=3i …(i)\\ 2z+... &3z+(2+i)w=11-i\\ &(2-i)z-w=-1+i\end{align} ====Solution==== Given that \begin{align}3z+\left( 2+i \righ
- Question 9, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- \operatorname{Re}\left( z \right)\leq |z|$ ====Solution==== Given $z=3+2i$. Then $|z|=\sqrt{9+4}=\sqrt{13... \operatorname{Im}\left( z \right)\leq |z|$ ====Solution==== Given $z=3+2i$. Then $|z|=\sqrt{9+4}=\sqrt{13
- Question 8, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- rline{z}=2\operatorname{Re}\left( z \right)$. ====Solution==== Assume $z=a+ib$, then $\overline{z}=a-ib$. \... line{z}=2i\operatorname{Im}\left( z \right)$. ====Solution==== Assume that $z=a+ib$, then $\overline{z}=a-i... ratorname{Im}\left( z \right) \right]}^{2}}$. ====Solution==== Suppose $z=a+ib$, then $\overline{z}=a-ib$. ... that $z=\overline{z}\Rightarrow z$ is real. ====Solution==== Suppose $z=a+bi$ ... (1) Then $\overline{z
- Question 7, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- eal and imaginary parts $\dfrac{2+3i}{5-2i}$. ====Solution==== \begin{align}&\dfrac{2+3i}{5-2i} \\ =&\dfrac... $\dfrac{{{\left( 1+2i \right)}^{2}}}{1-3i}$. ====Solution==== \begin{align}&\dfrac{(1+2i)^2}{1-3i}\\ =&\df... ts $\dfrac{1-i}{{{\left( 1+i \right)}^{2}}}$. ====Solution==== \begin{align}&\dfrac{1-i}{{{\left( 1+i \righ... ginary parts ${{\left( 2a-bi \right)}^{-2}}$. ====Solution==== \begin{align}&{{\left( 2a-bi \right)}^{-2}}\
- Question 6, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- |{{z}_{1}}{{z}_{2}}|=|{{z}_{1}}||{{z}_{2}}|$. ====Solution==== Suppose ${{z}_{1}}=a+bi$ and ${{z}_{2}}=c+di... {1}}|}{|{{z}_{2}}|}$, where ${{z}_{2}}\ne 0$ ====Solution==== Suppose $z=a+bi$, then $|z|=\sqrt{a^2+b^2}$.
- Question 5, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- }=\overline{{{z}_{1}}}+\overline{{{z}_{2}}}$. ====Solution==== Given ${{z}_{1}}=2+4i$ and ${{z}_{2}}=1-3i$.... }}=\overline{{{z}_{1}}}\overline{{{z}_{2}}}$. ====Solution==== Given ${{z}_{1}}=2+3i$ and ${{z}_{2}}=2-3i$\... \overline{{{z}_{1}}}}{\overline{{{z}_{2}}}}$. ====Solution==== Given ${{z}_{1}}=-a-3bi$ and ${{z}_{2}}=2a-3
- Question 3 & 4, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- property w.r.t. addition and multiplication. ====Solution==== ${{z}_{1}}=\sqrt{3}+\sqrt{2}i$ ${{z}_{2}}=\s... icative inverse of the complex number $5+2i$. ====Solution==== Given $z=5+2i$. Here $a=5$ and $b=2$. Addit... of the complex number $\left( 7,-9 \right)$. ====Solution==== Given $z=(7,-9)=7-9i$. Here $a=7$ and $b=-9$.
- Question 2, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- property w.r.t. addition and multiplication. ====Solution==== Given ${{z}_{1}}=-1+i$, ${{z}_{2}}=3-2i$ and
- Question 1, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- property w.r.t. addition and multiplication. ====Solution==== Given ${{z}_{1}}=2+i$, ${{z}_{2}}=1-i$. First
- Question 11, Exercise 1.1 @fsc-part1-kpk:sol:unit01
- 1}}{{z}_{2}}}{\overline{{{z}_{1}}}} \right)$. ====Solution==== Given $z_1=2-i$ and $z_2=-2+i$, then $\overli... {1}{{{z}_{1}}\overline{{{z}_{1}}}} \right)$. ====Solution==== Given $z_1=2-i$, then $\overline{z_1}=2+i$. \
- Question 9 & 10, Exercise 1.1 @fsc-part1-kpk:sol:unit01
- ht)}{\left( 1+2i \right)\left( 2-i \right)}$. ====Solution==== Let \begin{align}z&=\dfrac{\left( 3-2i \right... ( \dfrac{1}{i} \right)}^{25}} \right]}^{3}}$. ====Solution==== \begin{align}i^{18}+\left(\dfrac{1}{i}\right)
- Question 8, Exercise 1.1 @fsc-part1-kpk:sol:unit01
- {4-i}{3+2i}$ in the standard form of $a+ib.$ ====Solution==== \begin{align}&\dfrac{1-2i}{2+i}+\dfrac{4-i}{3... \sqrt{-16}}$ in the standard form of $a+ib.$ ====Solution==== \begin{align}\dfrac{2+\sqrt{-9}}{-5-\sqrt{-16... {2}}}{4+3i}$ in the standard form of $a+ib.$ ====Solution==== \begin{align}\dfrac{\left( 1+i \right)\left(
- Question 7, Exercise 1.1 @fsc-part1-kpk:sol:unit01
- {2}}=2+3i$, evaluate $|{{z}_{1}}+{{z}_{2}}|$. ====Solution==== We know that $z_1=1+2i$ and $z_2=2+3i$, then ... _{2}}=2+3i$, evaluate $|{{z}_{1}}{{z}_{2}}|$. ====Solution==== We know that $z_1=1+2i$ and $z_2=2+3i$, then ... i$, evaluate $\left|\dfrac{z_1}{z_2}\right|$. ====Solution==== We know that $z_1=1+2i$ and $z_2=2+3i$, then