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- Question 1, Exercise 1.3
- linear equation with complex coefficient. \begin{align}&z-4w=3i\\ &2z+3w=11-5i\end{align} ====Solution==== Given that \begin{align}z-4w&=3i …(i)\\ 2z+3w&=11-5i …(ii)\end{align} Multiply $2$ by (i), we get\\ \begin{align}2z-8w&=6i …(i
- Question 5, Exercise 1.2
- -4i$ and $\overline{{{z}_{2}}}=1+3i$. Now \begin{align}z_1+z_2&=2+4i+1-3i\\ &=3+i \end{align} Now \begin{align}\overline{z_1+z_2}=3-i \ldots (1)\end{align} and \begin{align} \overline{z_1}+\overline{z_2}&=2-4i+1+3i\
- Question 6, Exercise 1.3
- z}^{4}}+{{z}^{2}}+1=0$\\ ====Solution==== \begin{align}{{z}^{4}}+{{z}^{2}}+1&=0\\ {{z}^{4}}+2\left( \dfr... 2}}+\dfrac{1}{2} \right)}^{2}}&=-\dfrac{3}{4}\end{align}\\ Take square root on both sides.\\ \begin{align}\left( {{z}^{2}}+\dfrac{1}{2} \right)&=\pm \sqrt{-\dfr... \dfrac{\sqrt{3}}{2}i \right)}^{\dfrac{1}{2}}}\end{align} =====Question 6(ii)===== Find the solutions of
- Question 2, Exercise 1.2
- is, $$(z_1+z_2)+z_3=z_1+(z_2+z_3).$$ Take \begin{align} {{z}_{1}}+{{z}_{2}}&=\left( -1+i \right)+\left( 3-2i \right)\\ &=2-i\end{align} So \begin{align} \left( {{z}_{1}}+{{z}_{2}} \right)+{{z}_{3}}&=\left( 2-i \right)+\left( 2-2i \right)\\ &=4-3i \ldots (1)\end{align} Now \begin{align} {{z}_{2}}+{{z}_{3}}&=\left( 3-
- Question 8, Exercise 1.2
- Assume $z=a+ib$, then $\overline{z}=a-ib$. \begin{align}z+\overline{z}&=\left( a+ib \right)+\left( a-ib \... erline{z}&=2\operatorname{Re}\left( z \right)\end{align} =====Question 8(ii)===== Show that $z-\overlin... e that $z=a+ib$, then $\overline{z}=a-ib$. \begin{align}z-\overline{z}&=\left( a+ib \right)-\left( a-ib \... =2bi\\ z-\overline{z}&=2i\operatorname{Im}(z)\end{align} =====Question 8(iii)===== Show that $z\overlin
- Question 7, Exercise 1.1
- know that $z_1=1+2i$ and $z_2=2+3i$, then \begin{align} {{z}_{1}}+{{z}_{2}}&=1+2i+2+3i\\ &=1+2+2i+3i\\ &=3+5i \end{align} Now \begin{align} |z_1+z_2|&=\sqrt{3^2+5^2}\\ &=\sqrt{9+25}\\ &=\sqrt{34}\end{align} =====Question 7(ii)===== If ${{z}_{1}}=1+2i$
- Question 7, Exercise 1.2
- ts $\dfrac{2+3i}{5-2i}$. ====Solution==== \begin{align}&\dfrac{2+3i}{5-2i} \\ =&\dfrac{2+3i}{5-2i}\times... 4+19i}{29}\\ =&\dfrac{4}{29}+\dfrac{19}{29}i \end{align} Real part $=\dfrac{4}{29}$\\ Imaginary part $=\d... i \right)}^{2}}}{1-3i}$. ====Solution==== \begin{align}&\dfrac{(1+2i)^2}{1-3i}\\ =&\dfrac{1-4+4i}{1-3i}\... {-15-5i}{10}\\ =&\dfrac{-3}{2}-\dfrac{1}{2}i\end{align} Real part $=\dfrac{-3}{2}$ \\ Imaginary part $=
- Question 2, Exercise 1.3
- factor of $P(z)$ iff $P(a)=0$. Put $z=-2$ \begin{align} P(-2)&=(-2)^3+6(-2)+20\\ &=-8-12+20\\ &=0\end{align} Thus $z+2$ is a factor of ${{z}^{3}}+6z+20$.\\ By... & -2 & 10 & 0 \\ \end{array}$$ This gives \begin{align} P(z)&=(z+2)(z^2-2z+10)\\ &=(z+2)\left(z^2-2z+1+... -1)^2-(3i)^2\right]\\ &=(z+2)(z-1+3i)(z-1-3i)\end{align} =====Question 2(ii)===== Factorize the polynomi
- Question 2 & 3, Exercise 1.1
- i}^{122}}+{{i}^{153}}=0$. ====Solution==== \begin{align}L.H.S.&={{i}^{107}}+{{i}^{112}}+{{i}^{122}}+{{i}^... ( -1 \right)}^{76}}\\ &=-i+1-1+i\\ &=0=R.H.S.\end{align} =====Question 3(i)===== Add the complex numbe... ),-2\left( 1-3i \right)$. ====Solution==== \begin{align}& 3\left( 1+2i \right)+-2\left( 1-3i \right)\\ &=... ft( 3-2 \right)+\left( 6+6 \right)i\\ &=1+12i\end{align} =====Question 3(ii)===== Add the complex numbe
- Question 6, Exercise 1.1
- nswer in the form $a+ib$. ====Solution==== \begin{align}\dfrac{4+i}{3+5i}&=\dfrac{4+i}{3+5i}\times \dfrac... dfrac{17}{34}i\\ &=\dfrac{1}{2}-\dfrac{1}{2}i\end{align} =====Question 6(ii)===== Perform the indicate... nswer in the form $a+ib$. ====Solution==== \begin{align}\dfrac{1}{-8+i}&=\dfrac{1}{-8+i}\times \dfrac{-8-... -8-i}{64+1}\\ &=\dfrac{-8}{65}-\dfrac{1}{65}i\end{align} =====Question 6(iii)===== Perform the indicate
- Question 9 & 10, Exercise 1.1
- \left( 2-i \right)}$. ====Solution==== Let \begin{align}z&=\dfrac{\left( 3-2i \right)\left( 2+3i \right)}... {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}\\ &=\dfr... 3+16i}{25}\\ &=\dfrac{63}{25}+\dfrac{16}{25}i\end{align} =====Question 10===== Evalute ${{\left[ {{i}
- Question 6, Exercise 1.2
- +b^2}|$ and $|z_2=\sqrt{c^2+d^2}|$.\\ Now \begin{align} L.H.S.&=|{{z}_{1}}{{z}_{2}}|\\ &=|(a+bi)(c+di)|\... {{d}^{2}}}\\ &=|{{z}_{1}}||{{z}_{2}}|=R.H.S. \end{align} **Alternative Method**\\ We know $|z|^2=z\bar{z}$, so we have \begin{align}|{{z}_{1}}{{z}_{2}}{{|}^{2}}&={{z}_{1}}{{z}_{2}}\... ,|{{z}_{1}}{{z}_{2}}|&=|{{z}_{1}}||{{z}_{2}}|\end{align} proved. =====Question 6(ii)===== Show that for
- Question 5, Exercise 1.3
- \,b=1$ and $c=3$\\ Quadratic formula is\\ \begin{align}z&=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\ z&=\... {11}}{2}i,-\dfrac{1}{2}-\dfrac{\sqrt{11}}{2}i\end{align} =====Question 5(ii)===== Find the solutions of ... b=-1$ and $c=-1$\\ Quadratic formula is\\ \begin{align}z&=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\ z&=\... &=\dfrac{1+\sqrt{5}}{2},\dfrac{1-\sqrt{5}}{2}\end{align} =====Question 5(iii)===== Find the solutions of
- Question 2 & 3, Review Exercise 1
- =0$, $\forall n\in N$ \\ ====Solution==== \begin{align}{{i}^{n}}+{{i}^{n+1}}+{{i}^{n+2}}+{{i}^{n+3}}&=0\... i \right)\\ &=i\left( 0 \right)\\ &=0=R.H.S.\end{align} =====Question 3(i)===== Express the complex nu... in the form of $x+iy$.\\ ====Solution==== \begin{align}\left( 1+3i \right)-\left( 5+7i \right)&=1+3i-5-7i\\ &=1-5+3i-7i\\ &=-4-4i\end{align} =====Question 3(iii)===== Express the complex n
- Question 4 & 5, Review Exercise 1
- }_{1}}-{{z}_{2}}+1}|$.\\ ====Solution==== \begin{align}{{z}_{1}}&=2-i,\\ {{z}_{2}}&=1+i,\\ \dfrac{{{z}_{... \left( 1+i \right)}{2}\\ &=\left( 1+i \right)\end{align} Now\\ \begin{align}|\dfrac{{{z}_{1}}+{{z}_{2}}+1}{{{z}_{1}}-{{z}_{2}}+1}|&=\sqrt{{{1}^{2}}+{{1}^{2}}}\\ &=\sqrt{2}\end{align} =====Question 5===== Find the modulus of $\dfr