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- Question 2 & 3, Exercise 1.1
- +{{i}^{112}}+{{i}^{122}}+{{i}^{153}}=0$. GOOD ====Solution==== \begin{align}L.H.S.&={{i}^{107}}+{{i}^{112}}+... $3\left( 1+2i \right),-2\left( 1-3i \right)$. ====Solution==== \begin{align}& 3\left( 1+2i \right)+-2\left( ... 2}-\dfrac{2}{3}i,\dfrac{1}{4}-\dfrac{1}{3}i$. ====Solution==== \begin{align}&\left( \dfrac{1}{2}-\dfrac{2}{3... sqrt{2},1 \right),\left( 1,\sqrt{2} \right)$. ====Solution==== \begin{align}&\left( \sqrt{2},1 \right)+\left
- Question 1, Exercise 1.1
- i)===== Simplify ${{i}^{9}}+{{i}^{19}}$. GOOD ====Solution==== \begin{align}{{i}^{9}}+{{i}^{19}}&=i\cdot{{i}... = Simplify ${{\left( -i \right)}^{23}}$. GOOD ====Solution==== \begin{align}{{\left( -i \right)}^{23}}&={{\l... ${{\left( -1 \right)}^{\frac{-23}{2}}}$. GOOD ====Solution==== \begin{align}{{\left( -1 \right)}^{\frac{-23}... ${{\left( -1 \right)}^{\frac{15}{2}}}$. GOOD ====Solution==== \begin{align}{{\left( -1 \right)}^{\frac{15}{
- Question 6, 7 & 8, Review Exercise 1
- 6===== Find conjugate of $\dfrac{1}{3+4i}$.\\ ====Solution==== Suppose $$z=\dfrac{1}{3+4i}.$$ Then \begin{a... iplicative inverse of $\dfrac{3i+2}{3-2i}$.\\ ====Solution==== \begin{align}\dfrac{3i+2}{3-2i}\\ \dfrac{3i+2... the quadrative equation $z+\dfrac{2}{z}=2.$\\ ====Solution==== \begin{align}z+\dfrac{2}{z}&=2\\ {{z}^{2}}+2&
- Question 4 & 5, Review Exercise 1
- +{{z}_{2}}+1}{{{z}_{1}}-{{z}_{2}}+1}\right|$. ====Solution==== Given $z_1=2-i$ and $z_2=1+i$, so we have \be... lus of $\dfrac{1+i}{1-i}-\dfrac{1-i}{1+i}$.\\ ====Solution==== \begin{align}\dfrac{1+i}{1-i}-\dfrac{1-i}{1+i
- Question 2 & 3, Review Exercise 1
- i}^{n+2}}+{{i}^{n+3}}=0$, $\forall n\in N$ \\ ====Solution==== \begin{align}{{i}^{n}}+{{i}^{n+1}}+{{i}^{n+2... left( 5+7i \right)$ in the form of $x+iy$.\\ ====Solution==== $\left( 1+3i \right)+\left( 5+7i \right)=1+5... left( 5+7i \right)$ in the form of $x+iy$.\\ ====Solution==== \begin{align}\left( 1+3i \right)-\left( 5+7i... left( 5+7i \right)$ in the form of $x+iy$.\\ ====Solution==== \begin{align}(1+3i)(5+7i)&=5+7i+15i+21{{i}^{
- Question 6, Exercise 1.3
- ns of the equation ${{z}^{4}}+{{z}^{2}}+1=0$. ====Solution==== $$z^4+z^2+1=0$$ $$z^4+2z^2+1-z^2=0$$ $$( z^... the solutions of the equation ${{z}^{3}}=-8$. ====Solution==== Given: $$z^3=-8.$$ This gives \begin{align} ... e equation ${{\left( z-1 \right)}^{3}}=-1$.\\ ====Solution==== Given: $$(z-1)^3=-1.$$ Since we have $$(a-b... e solutions of the equation ${{z}^{3}}=1$. \\ ====Solution==== Given $$z^3=1.$$ Thus, we have \begin{align}
- Question 5, Exercise 1.3
- solutions of the equation ${{z}^{2}}+z+3=0$. ====Solution==== Given: $${{z}^{2}}+z+3=0.$$ According to the... solutions of the equation ${{z}^{2}}-1=z$.\\ ====Solution==== Given: $${{z}^{2}}-1=z$$ $$\implies {{z}^{2}... lutions of the equation ${{z}^{2}}-2z+i=0$\\ ====Solution==== Given: $${{z}^{2}}-2z+i=0$$ According to the... e solutions of the equation ${{z}^{2}}+4=0$\\ ====Solution==== Given: $${{z}^{2}}+4=0$$ \begin{align} \impl
- Question 3 & 4, Exercise 1.3
- satisfied the equation ${{z}^{2}}+2z+2=0$\\ ====Solution==== Given: $$z^2+2z_1+2=0\quad \ldots (i)$$ Put ... ====Question 4===== Determine weather $1+2i$ is a solution of ${{z}^{2}}-2z+5=0$\\ ====Solution==== Given: $$z^2-2z+5=0 \ldots (i)$$ Put $z=1+2i$ in equaiton (i), we... \ &=1-4+5\\ &=0\end{align} This implies $1+2i$ is solution of the given equation. ====Go To==== <text align=
- Question 2, Exercise 1.3
- actors. $$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 9, Exercise 1.2
- \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 $|z|=\sqrt{9+4}=\sqrt{13
- Question 1, Exercise 1.3
- in{align}&z-4w=3i\\ &2z+3w=11-5i\end{align} ====Solution==== Given that \begin{align}z-4w&=3i …(i)\\ 2z... t. \begin{align}&z+w=3i\\ &2z+3w=2\end{align} ====Solution==== Given that \begin{align}z+w&=3i …(i)\\ 2z+... &3z+(2+i)w=11-i\\ &(2-i)z-w=-1+i\end{align} ====Solution==== Given that \begin{align}3z+\left( 2+i \righ
- Question 8, Exercise 1.2
- rline{z}=2\operatorname{Re}\left( z \right)$. ====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
- 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
- |{{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
- }=\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