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- 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 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 $\overline{z
- Question 2 & 3, Review Exercise 1
- i}^{n+2}}+{{i}^{n+3}}=0$, $\forall n\in N$ \\ ====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}\left( 1+3i \right)\left( 5+7i
- Question 1, Exercise 1.1
- on 1(i)===== Simplify ${{i}^{9}}+{{i}^{19}}$. ====Solution==== \begin{align}{{i}^{9}}+{{i}^{19}}&=i\cdot{{i}... )===== Simplify ${{\left( -i \right)}^{23}}$. ====Solution==== \begin{align}{{\left( -i \right)}^{23}}&={{\l... lify ${{\left( -1 \right)}^{\frac{-23}{2}}}$. ====Solution==== \begin{align}{{\left( -1 \right)}^{\frac{-23}... plify ${{\left( -1 \right)}^{\frac{15}{2}}}$. ====Solution==== \begin{align}{{\left( -1 \right)}^{\frac{15}{
- Question 2 & 3, Exercise 1.1
- 107}}+{{i}^{112}}+{{i}^{122}}+{{i}^{153}}=0$. ====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 6, Exercise 1.1
- i}$ and write the answer in the form $a+ib$. ====Solution==== \begin{align}\dfrac{4+i}{3+5i}&=\dfrac{4+i}{3... i}$ and write the answer in the form $a+ib$. ====Solution==== \begin{align}\dfrac{1}{-8+i}&=\dfrac{1}{-8+i}... i}$ and write the answer in the form $a+ib$. ====Solution==== \begin{align}\dfrac{1}{7-3i}&=\dfrac{1}{7-3i}... i}$ and write the answer in the form $a+ib$. ====Solution==== \begin{align}\dfrac{6+i}{i}&=\dfrac{6+i}{i}\t
- 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 5, Exercise 1.3
- solutions of the equation ${{z}^{2}}+z+3=0$\\ ====Solution==== ${{z}^{2}}+z+3=0$\\ According to the quadrat... solutions of the equation ${{z}^{2}}-1=z$.\\ ====Solution==== ${{z}^{2}}-1=z$\\ ${{z}^{2}}-z-1=0$\\ Accord... lutions of the equation ${{z}^{2}}-2z+i=0$\\ ====Solution==== ${{z}^{2}}-2z+i=0$\\ According to the quadra... e solutions of the equation ${{z}^{2}}+4=0$\\ ====Solution==== ${{z}^{2}}+4=0$\\ According to the quadratic
- Question 6, Exercise 1.3
- s of the equation ${{z}^{4}}+{{z}^{2}}+1=0$\\ ====Solution==== \begin{align}{{z}^{4}}+{{z}^{2}}+1&=0\\ {{z}... he solutions of the equation ${{z}^{3}}=-8$\\ ====Solution==== \begin{align}{{z}^{3}}&=-8\\ {{z}^{3}}+{{2}^... e equation ${{\left( z-1 \right)}^{3}}=-1$.\\ ====Solution==== \begin{align}{{\left( z-1 \right)}^{3}}&=-1\... he solutions of the equation ${{z}^{3}}=1$ \\ ====Solution==== \begin{align}{{z}^{3}}&=1\\ {{z}^{3}}-{{1}^{
- Question 4, Exercise 1.1
- irst $\left( a,0 \right)\left( 2,-b \right)$. ====Solution==== \begin{align}&\left( a,0 \right)-\left( 2,-b ... {1}{2} \right)\left( 3,\dfrac{1}{2} \right)$. ====Solution==== \begin{align}&\left( -3,\dfrac{1}{2} \right)-... t $3\sqrt{3}-5\sqrt{7}i,\sqrt{3}+2\sqrt{7}i$. ====Solution==== \begin{align}&\left(3\sqrt{3}-5\sqrt{7}i \rig
- Question 5, Exercise 1.1
- == Multiply the complex number $8i+11,-7+5i$. ====Solution==== \begin{align}&(8i+11)\times (-7+5i)\\ &=\left... the complex number $3i,2\left( 1-i \right)$. ====Solution==== \begin{align}&3i\times 2\left( 1-i \right)\\ ... ber $\sqrt{2}+\sqrt{3i},2\sqrt{2}-\sqrt{3i}$. ====Solution==== \begin{align}&\left( \sqrt{2}+\sqrt{3}i \righ
- Question 7, Exercise 1.1
- {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
- Question 8, Exercise 1.1
- {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 3 & 4, Exercise 1.2
- 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 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