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- Question 7, Exercise 1.2
- estion 7(i)===== Separate into real and imaginary parts $\dfrac{2+3i}{5-2i}$. ====Solution==== \begin{align}&\dfrac{2+3i}{... =&\dfrac{4}{29}+\dfrac{19}{29}i \end{align} Real part $=\dfrac{4}{29}$\\ Imaginary part $=\dfrac{19}{29}$ =====Question 7(ii)===== Separate into real and imaginary parts $\dfrac{{{\left( 1+2i \right)}^{2}}}{1-3i}$. ====Solution==== \begin{align}&\dfrac
- Question 6, Exercise 1.2
- ==== <text align="left"><btn type="primary">[[fsc-part1-kpk:sol:unit01:ex1-2-p4|< Question 5]]</btn></text> <text align="right"><btn type="success">[[fsc-part1-kpk:sol:unit01:ex1-2-p6|Question 7 >]]</btn></text>
- Question 3 & 4, Exercise 1.2
- ==== <text align="left"><btn type="primary">[[fsc-part1-kpk:sol:unit01:ex1-2-p2|< Question 2]]</btn></text> <text align="right"><btn type="success">[[fsc-part1-kpk:sol:unit01:ex1-2-p4|Question 5 >]]</btn></text>
- Question 5, Exercise 1.2
- ==== <text align="left"><btn type="primary">[[fsc-part1-kpk:sol:unit01:ex1-2-p3|< Question 3 & 4]]</btn></text> <text align="right"><btn type="success">[[fsc-part1-kpk:sol:unit01:ex1-2-p5|Question 6 >]]</btn></text>
- Question 6, Exercise 1.3
- nd the solutions of the equation ${{z}^{4}}+{{z}^{2}}+1=0$\\ ====Solution==== \begin{align}{{z}^{4}}+{{z}^{2}}+1&=0\\ {{z}^{4}}+2\left( \dfrac{1}{2} \right){{z}^{2}}+\dfrac{1}{4}-\dfrac{1}{4}+1&=0\\ {{\left( {{z}^{2}}+\dfrac{1}{2}
- Question 2, Exercise 1.3
- ====== Question 2, Exercise 1.3 ====== Solutions of Question 2 of Exercise 1.3 of Unit 01: Complex Numbers. This is un... KPTB or KPTBB) Peshawar, Pakistan. =====Question 2(i)===== Factorize the polynomial $P(z)$ into linear factors. $$P\left( z \right)={{z}^{3}}+6z+20$$ ====Solution==== Given: $$p\left( z \right)=
- Question 2, Exercise 1.2
- ==== <text align="left"><btn type="primary">[[fsc-part1-kpk:sol:unit01:ex1-2-p1|< Question 1]]</btn></text> <text align="right"><btn type="success">[[fsc-part1-kpk:sol:unit01:ex1-2-p3|Question 3, 4 >]]</btn></text>
- Question 5, Exercise 1.3
- i)===== Find the solutions of the equation ${{z}^{2}}+z+3=0$\\ ====Solution==== ${{z}^{2}}+z+3=0$\\ According to the quadratic formula, we have\\ $a=1,... ula is\\ \begin{align}z&=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\ z&=\dfrac{-\left( 1 \right)\pm \sqrt{{{\left( 1 \right)}^{2}}-4\left( 1 \right)\left(
- Question 7, Exercise 1.1
- Pakistan. =====Question 7(i)===== If ${{z}_{1}}=1+2i$ and ${{z}_{2}}=2+3i$, evaluate $|{{z}_{1}}+{{z}_{2}}|$. ====Solution==== We know that $z_1=1+2i$ and $z_2=2+3i$, then
- Question 1, Exercise 1.3
- =1, \quad w=3-2i.$$ ====Go to ==== <text align="left"><btn type="success">[[fsc-part1-kpk:sol:unit01:ex1-3-p2|Question 2 >]]</btn></text>
- Question 3 & 4, Exercise 1.3
- ==== <text align="left"><btn type="primary">[[fsc-part1-kpk:sol:unit01:ex1-3-p2 |< Question 2]]</btn></text> <text align="right"><btn type="success">[[fsc-part1-kpk:sol:unit01:ex1-3-p4|Question 5 >]]</btn></te
- Question 6, 7 & 8, Review Exercise 1
- 4i}{3-4i}\\ &=\dfrac{3-4i}{9+16}\\ &=\dfrac{3-4i}{25}\\ &=\dfrac{3}{25}-\dfrac{4i}{25}\end{align} =====Question 7===== Find the multiplicative inverse of $\dfrac{3i+2}{3-2i}$.\\ ====Solution==== \begin{align}\dfrac{3
- Question 4 & 5, Review Exercise 1
- ==== <text align="left"><btn type="primary">[[fsc-part1-kpk:sol:unit01:Review-ex-1-p2 |< Question 2 & 3]]</btn></text> <text align="right"><btn type="success">[[fsc-part1-kpk:sol:unit01:Review-ex-1-p4|Question 6 & 7 >]]
- 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 Numbers. Th... KPTB or KPTBB) Peshawar, Pakistan. =====Question 2===== Prove that ${{i}^{107}}+{{i}^{112}}+{{i}^{122}}+{{i}^{153}}=0$. ====Solution==== \begin{align}L
- Question 8, Exercise 1.1
- an. =====Question 8(i)===== Express the $\dfrac{1-2i}{2+i}+\dfrac{4-i}{3+2i}$ in the standard form of $a+ib.$ ====Solution==== \begin{align}&\dfrac{1-2i}{2+i}+\dfrac{4-i}{3+2i}\\ &=\dfrac{\left( 3+2i \