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- Question 6, Exercise 1.3
- +{{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} \right)}^{2}}+\dfrac{4-1}{4}&=0\\ {{\left( {{z}^{2}}+\dfrac{1}{2} \right)}^{2}}+\dfrac{3}{4}=0\\ {{\left( {{z}^{2}}+\dfrac{1}{2} \right)}^{2}}&=-\dfrac{3}{4}\end{align}\\ Take square ro
- Question 3 & 4, Exercise 1.2
- \begin{align}{{z}_{1}}\left( {{z}_{2}}+{{z}_{3}} \right)&={{z}_{1}}{{z}_{2}}+{{z}_{1}}{{z}_{3}}\\ {{z}_{... }&=\sqrt{2}-\sqrt{3}i+2+3i\\ &=\left( \sqrt{2}+2 \right)+\left( 3-\sqrt{3} \right)i\\ L.H.S.&={{z}_{1}}\left( {{z}_{2}}+{{z}_{3}} \right)\\ &=\left( \sqrt{3}+\sqrt{2}i \right)\left( \lef
- Question 2 & 3, Exercise 1.1
- +i^{122}+i\cdot i^{152}\\ &=i.{{\left( {{i}^{2}} \right)}^{53}}+{{\left( {{i}^{2}} \right)}^{56}}+{{\left( {{i}^{2}} \right)}^{61}}+i.{{\left( {{i}^{2}} \right)}^{76}}\\ &=i.{{\left( -1 \right)}^{53}}+{{\left( -1 \right)}^{56}}+
- Question 2 & 3, Review Exercise 1
- {3}}\\ &={{i}^{n}}\left( 1+i+{{i}^{2}}+{{i}^{3}} \right)\\ &={{i}^{n}}\left( 1+i+{{i}^{2}}+{{i}^{3}} \right)\\ &=i\left( 1+i+{{\left( i \right)}^{2}}+i{{\left( i \right)}^{2}} \right)\\ &=i\left( 1+i+\left( -1 \right)+i\left( -1 \right) \right)\\
- Question 9 & 10, Exercise 1.1
- ==== Find the conjugate of $\dfrac{\left( 3-2i \right)\left( 2+3i \right)}{\left( 1+2i \right)\left( 2-i \right)}$. ====Solution==== Let \begin{align}z&=\dfrac{\left( 3-2i \right)\left( 2+3i \right)
- Question 7, Exercise 1.2
- o real and imaginary parts $\dfrac{{{\left( 1+2i \right)}^{2}}}{1-3i}$. ====Solution==== \begin{align}&\... al and imaginary parts $\dfrac{1-i}{{{\left( 1+i \right)}^{2}}}$. ====Solution==== \begin{align}&\dfrac{1-i}{{{\left( 1+i \right)}^{2}}}\\ =&\dfrac{1-i}{1-1+2i}\\ =&\dfrac{1-i}{2... te into real and imaginary parts ${{\left( 2a-bi \right)}^{-2}}$. ====Solution==== \begin{align}&{{\left
- Question 1, Exercise 1.1
- }}+i\cdot{{i}^{18}}\\ &=i\cdot{{\left( {{i}^{2}} \right)}^{4}}+i\cdot{{\left( {{i}^{2}} \right)}^{9}}\\ &=i\cdot{{\left( -1 \right)}^{4}}+i\cdot{{\left( -1 \right)}^{9}}\\ &=i\cdot1+i\cdot\left( -1 \right)\\ &=i-i\\ &=0\end{align} =
- Question 8, Exercise 1.2
- that $z+\overline{z}=2\operatorname{Re}\left( z \right)$. ====Solution==== Assume $z=a+ib$, then $\over... =a-ib$. \begin{align}z+\overline{z}&=\left( a+ib \right)+\left( a-ib \right)\\ &=a+ib+a-ib\\ &=2a\\ z+\overline{z}&=2\operatorname{Re}\left( z \right)\end{align} =====Question 8(ii)===== Show that
- Question 4, Exercise 1.1
- the second complex number from first $\left( a,0 \right)\left( 2,-b \right)$. ====Solution==== \begin{align}&\left( a,0 \right)-\left( 2,-b \right)\\ &=\left( a+0i \right)-\left( 2-bi \right)\\ &=\left( a-2 \right)+\left( 0+b \righ
- Question 5, Exercise 1.1
- in{align}&(8i+11)\times (-7+5i)\\ &=\left( 11+8i \right)\times \left( -7+5i \right)\\ &=\left( -77+40{{i}^{2}} \right)+\left( 55-56 \right)i\\ &=\left( -77+40\left( -1 \right) \right)+\left( 55-56 \right)i\\ &=\left( -77-4
- Question 5, Exercise 1.3
- \sqrt{{{b}^{2}}-4ac}}{2a}\\ z&=\dfrac{-\left( 1 \right)\pm \sqrt{{{\left( 1 \right)}^{2}}-4\left( 1 \right)\left( 3 \right)}}{2\left( 1 \right)}\\ z&=\dfrac{-1\pm \sqrt{1-12}}{2}\\ z&=\dfrac{-1\pm \sqrt{11}}{2}i
- Question 8, Exercise 1.1
- i}{2+i}+\dfrac{4-i}{3+2i}\\ &=\dfrac{\left( 3+2i \right)\left( 1-2i \right)+\left( 2+i \right)\left( 4-i \right)}{\left( 2+i \right)\left( 3+2i \right)}\\ &=\dfrac{\left( 3+4+2i-6i \right)+\left( 8+
- Question 6, Exercise 1.2
- r{z}_2\\ &=\left( {{z}_{1}}\overline{{{z}_{1}}} \right)\left( {{z}_{2}}\overline{{{z}_{2}}} \right)\\ &=|{{z}_{1}}{{|}^{2}}|{{z}_{2}}{{|}^{2}}\\ \Rightarrow \,\,\,\,|{{z}_{1}}{{z}_{2}}|&=|{{z}_{1}}||{{z... }_{2}}$that $\left| \dfrac{{{z}_{1}}}{{{z}_{2}}} \right|=\dfrac{|{{z}_{1}}|}{|{{z}_{2}}|}$, where ${{z}_{
- Question 2, Exercise 1.3
- lynomial $P(z)$ into linear factors. $$P\left( z \right)={{z}^{3}}+6z+20$$ ====Solution==== Given: $$p\left( z \right)={{z}^{3}}+6z+20$$ By factor theorem, $(z-a)$ is ... P(z)&=(z+2)(z^2-2z+10)\\ &=(z+2)\left(z^2-2z+1+9\right)\\ &=(z+2)\left[(z-1)^2-(3i)^2\right]\\ &=(z+2)(z-1+3i)(z-1-3i)\end{align} =====Question 2(ii)===== Fac
- Question 7, Exercise 1.1
- en \begin{align} {{z}_{1}}{{z}_{2}}&=\left( 1+2i \right)\times \left( 2+3i \right)\\ &=\left( 2-6 \right)+\left( 3+4 \right)i\\ &=-4+7i. \end{align} Now \begin{align} |z_1 z_2|&=\sqrt{(-4)^2+7^2}\\ &=\sqrt{16+