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- Question 2 & 3, Exercise 1.1
- ====== Question 2 & 3, Exercise 1.1 ====== Solutions of Question 2 & 3 of Exercise 1.1 of Unit 01: Complex N... +{{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
- Question 1, Exercise 1.1
- ====== Question 1, Exercise 1.1 ====== Solutions of Question 1 of Exercise 1.1 of Unit 01: Complex Numbers. ... 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}
- Question 6, 7 & 8, Review Exercise 1
- ===== Question 6, 7 & 8, Review Exercise 1 ====== Solutions of Question 6, 7 & 8 of Review Exercise 1 of Uni... 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
- ====== Question 4 & 5, Review Exercise 1 ====== Solutions of Question 4 & 5 of Review Exercise 1 of Unit 01:... +{{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
- ====== Question 2 & 3, Review Exercise 1 ====== Solutions of Question 2 & 3 of Review Exercise 1 of Unit 01:... 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
- Question 1, Review Exercise 1
- ====== Question 1, Review Exercise 1 ====== Solutions of Question 1 of Review Exercise 1 of Unit 01: Complex
- Question 6, Exercise 1.3
- ====== Question 6, Exercise 1.3 ====== Solutions of Question 6 of Exercise 1.3 of Unit 01: Complex Numbers. ... awar, Pakistan. =====Question 6(i)===== Find the solutions of the equation ${{z}^{4}}+{{z}^{2}}+1=0$. ====Solution==== $$z^4+z^2+1=0$$ $$z^4+2z^2+1-z^2=0$$ $$( z^2... qrt{3}}{2}i$$ =====Question 6(ii)===== Find the solutions of the equation ${{z}^{3}}=-8$. ====Solution====
- Question 5, Exercise 1.3
- ====== Question 5, Exercise 1.3 ====== Solutions of Question 5 of Exercise 1.3 of Unit 01: Complex Numbers. ... awar, Pakistan. =====Question 5(i)===== Find the solutions of the equation ${{z}^{2}}+z+3=0$. ====Solution==== Given: $${{z}^{2}}+z+3=0.$$ According to the quadrati... &=\dfrac{-1\pm \sqrt{11}}{2}i\end{align} Thus the solutions of the given equation are $-\dfrac{1}{2}\pm\dfra
- Question 3 & 4, Exercise 1.3
- ====== Question 3 & 4, Exercise 1.3 ====== Solutions of Question 3 & 4 of Exercise 1.3 of Unit 01: Complex N... 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
- Question 2, Exercise 1.3
- ====== Question 2, Exercise 1.3 ====== Solutions of Question 2 of Exercise 1.3 of Unit 01: Complex Numbers. ... 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(
- Question 9, Exercise 1.2
- ====== Question 9, Exercise 1.2 ====== Solutions of Question 9 of Exercise 1.2 of Unit 01: Complex Numbers.... \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 $|z|=\sqrt{9+4}=\sqrt{13
- Question 1, Exercise 1.3
- ====== Question 1, Exercise 1.3 ====== Solutions of Question 1 of Exercise 1.3 of Unit 01: Complex Numbers. ... in{align}&z-4w=3i\\ &2z+3w=11-5i\end{align} ====Solution==== Given that \begin{align}z-4w&=3i …(i)\\ 2z... t. \begin{align}&z+w=3i\\ &2z+3w=2\end{align} ====Solution==== Given that \begin{align}z+w&=3i …(i)\\ 2z+... &3z+(2+i)w=11-i\\ &(2-i)z-w=-1+i\end{align} ====Solution==== Given that \begin{align}3z+\left( 2+i \righ
- Question 8, Exercise 1.2
- ====== Question 8, Exercise 1.2 ====== Solutions of Question 8 of Exercise 1.2 of Unit 01: Complex Numbers. ... rline{z}=2\operatorname{Re}\left( z \right)$. ====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 $\overline{z}=a-ib$.
- Question 7, Exercise 1.2
- ====== Question 7, Exercise 1.2 ====== Solutions of Question 7 of Exercise 1.2 of Unit 01: Complex Numbers. ... 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
- Question 6, Exercise 1.2
- ====== Question 6, Exercise 1.2 ====== Solutions of Question 6 of Exercise 1.2 of Unit 01: Complex Numbers. ... |{{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 $|z|=\sqrt{a^2+b^2}$.