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Question 2, Exercise 1.2
6 Hits, Last modified: 5 months ago
2 z_3&=(3-2i)\cdot(2-2i)\\ &=(6-4)+(-4-6)i\\ &=2-10i \end{align} So \begin{align} z_1(z_2 z_3)&=(-1+i)\cdot (2-10i)\\ &=(-2+10)+(2+10)i\\ &=8+12i \ldots (3)\end{align} Now, we take \begin{align} z_1 z_2 &=(-1+i)\cdot (3-2i)\\ &
Question 8, Exercise 1.1
4 Hits, Last modified: 5 months ago
ight)i}{16+49}\\ &=\dfrac{50-120i}{65}\\ &=\dfrac{10-24i}{13}\\ &=\dfrac{10}{13}-\dfrac{24i}{13}\end{align} =====Question 8(ii)===== Express the $\dfrac{... 4i}\times \dfrac{-5+4i}{-5+4i}\\ &=\dfrac{\left( -10-12 \right)+\left( 8-15 \right)i}{25+16}\\ &=\dfra... ign="right"><btn type="success">[[fsc-part1-kpk:sol:unit01:ex1-1-p8|Question 9 & 10 >]]</btn></text>
Question 2 & 3, Exercise 1.1
3 Hits, Last modified: 5 months ago
Pakistan. =====Question 2===== Prove that ${{i}^{107}}+{{i}^{112}}+{{i}^{122}}+{{i}^{153}}=0$. ====Solution==== \begin{align}L.H.S.&={{i}^{107}}+{{i}^{112}}+{{i}^{122}}+{{i}^{153}}\\ &=i\cdot i^{106}+i^{112}+i^{122}+i\cdot i^{152}\\ &=i.{{\left( {
Question 9 & 10, Exercise 1.1
3 Hits, Last modified: 5 months ago
====== Question 9 & 10, Exercise 1.1 ====== Solutions of Question 9 & 10 of Exercise 1.1 of Unit 01: Complex Numbers. This i... 3}{25}+\dfrac{16}{25}i\end{align} =====Question 10===== Evalute ${{\left[ {{i}^{18}}+{{\left( \dfr
Question 7, Exercise 1.2
3 Hits, Last modified: 5 months ago
}{5+2i} \quad \text{by rationalizing} \\ =&\dfrac{10-6+15i+4i}{25+4}\\ =&\dfrac{4+19i}{29}\\ =&\dfrac{... \\ =&\dfrac{-3-12+4i-9i}{1+9}\\ =&\dfrac{-15-5i}{10}\\ =&\dfrac{-3}{2}-\dfrac{1}{2}i\end{align} Rea... mes \dfrac{-5-12i}{-5-12i}\\ =&\dfrac{45-480+200i+108i}{25+144}\\ =&\dfrac{-435+308i}{169}\\ =&\dfrac{
Question 2, Exercise 1.3
2 Hits, Last modified: 5 months ago
\downarrow & -2 & 4 & -20 \\ \hline & 1 & -2 & 10 & 0 \\ \end{array}$$ This gives \begin{align} P(z)&=(z+2)(z^2-2z+10)\\ &=(z+2)\left(z^2-2z+1+9\right)\\ &=(z+2)\left
Question 11, Exercise 1.1
1 Hits, Last modified: 5 months ago
align="left"><btn type="primary">[[fsc-part1-kpk:sol:unit01:ex1-1-p8|< Question 9, 10]]</btn></text>
Question 1, Exercise 1.3
1 Hits, Last modified: 5 months ago
i}{2-i}\\ &=\dfrac{16-8i-2i-1}{4+1}\\ &=\dfrac{15-10i}{5}=3-2i\end{align} Hence $$z=1, \quad w=3-2i.$$
Question 2 & 3, Review Exercise 1
1 Hits, Last modified: 5 months ago
( 1+3i \right)+\left( 5+7i \right)=1+5+3i+7i$ $=6+10i$ =====Question 3(ii)===== Express the complex n