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- Question 9 & 10, Exercise 1.1
- ight)}\\ &=\dfrac{6+6+9i-4i}{2+2+4i-i}\\ &=\dfrac{12+5i}{4+3i}\end{align} Now \begin{align}\bar{z}&=\dfrac{12-5i}{4-3i}\\ &=\dfrac{12-5i}{4-3i}\times \dfrac{4+3i}{4+3i}\\ &=\dfrac{48+15+36i-20i}{16+9+12i-12i}\\ &=\dfrac{63+16i}{25}\\ &=\dfrac{63}{25}+\
- Question 7, Exercise 1.2
- 1+3i} \quad \text{by rationalizing}\\ =&\dfrac{-3-12+4i-9i}{1+9}\\ =&\dfrac{-15-5i}{10}\\ =&\dfrac{-... 5i}{2+3i} \right)}^{2}}\\ =&\dfrac{16-25-40i}{4-9+12i}\\ =&\dfrac{-9-40i}{-5+12i}\\ =&\dfrac{-9-40i}{-5+12i}\times \dfrac{-5-12i}{-5-12i}\\ =&\dfrac{45-480+200i+108i}{25+144}\\ =&\d
- Question 2 & 3, Exercise 1.1
- ===Question 2===== Prove that ${{i}^{107}}+{{i}^{112}}+{{i}^{122}}+{{i}^{153}}=0$. GOOD ====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( {{i}^{2}}
- Question 8, Exercise 1.1
- 7i}{4-7i}\\ &=\dfrac{\left( 64-14 \right)-\left( 112+8 \right)i}{16+49}\\ &=\dfrac{50-120i}{65}\\ &=\dfrac{10-24i}{13}\\ &=\dfrac{10}{13}-\dfrac{24i}{13}\... \times \dfrac{-5+4i}{-5+4i}\\ &=\dfrac{\left( -10-12 \right)+\left( 8-15 \right)i}{25+16}\\ &=\dfrac{-
- Question 2, Exercise 1.2
- 3)&=(-1+i)\cdot (2-10i)\\ &=(-2+10)+(2+10)i\\ &=8+12i \ldots (3)\end{align} Now, we take \begin{align}... _3&=(-1+5i)\cdot (2-2i)\\ &=(-2+10)+(10+2)i\\ &=8+12i \ldots (4)\end{align} From (3) and (4), we get t
- Question 2, Exercise 1.3
- =-2$ \begin{align} P(-2)&=(-2)^3+6(-2)+20\\ &=-8-12+20\\ &=0\end{align} Thus $z+2$ is a factor of ${... =2$\\ \begin{align}P(-2)&=(-2)^3+6(-2)+20\\ &=-8-12+20=0\end{align} Thus $z-2$ is a factor of ${{z}^
- Question 6, Exercise 1.3
- dfrac{2\pm \sqrt{4-16}}{2}\\ &=\dfrac{2\pm \sqrt{-12}}{2}\\ &=\dfrac{2\pm 2\sqrt{3}i}{2}\\ &=1\pm \sqr... ght)}}{2\left( 1 \right)}\\ &=\dfrac{3\pm \sqrt{9-12}}{2}\\ &=\dfrac{3\pm \sqrt{-3}}{2}\\ &=\dfrac{3\p
- Question 6, Exercise 1.1
- {3+5i}\times \dfrac{3-5i}{3-5i}\\ &=\dfrac{\left( 12+5 \right)+\left( 3-20 \right)i}{9-25{{i}^{2}}}\\
- Question 5, Exercise 1.3
- ht)}}{2\left( 1 \right)}\\ &=\dfrac{-1\pm \sqrt{1-12}}{2}\\ &=\dfrac{-1\pm \sqrt{11}}{2}i\end{align} T