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- Question 1, Exercise 1.3
- aneous 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 …(iii)\end{align} Subtract (iii) from (ii), we get\\ \[\begin{array}{cccc} 2z&-8w&=6i \\ \mathop+\limits_{-}
- Question 5, Exercise 1.2
- 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\\... z}_{1}}}=2-3i$ and $\overline{{{z}_{2}}}=2+3i$. \begin{align}z_1 z_2 &=(2+3i)(2-3i)\\ &=2^2-(3i)^2\\ &=1
- Question 2, Exercise 1.2
- , that 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... 2-2i \right)\\ &=4-3i \ldots (1)\end{align} Now \begin{align} {{z}_{2}}+{{z}_{3}}&=\left( 3-2i \right)+\left( 2-2i \right)\\ &=5-4i\end{align} So \begin{align} {{z}_{1}}+\left({{z}_{2}}+{{z}_{3}} \right
- Question 2, Exercise 1.3
- is a factor of $P(z)$ iff $P(a)=0$. Put $z=-2$ \begin{align} P(-2)&=(-2)^3+6(-2)+20\\ &=-8-12+20\\ &=0\... 6z+20$.\\ By using synthetic division, we have $$\begin{array}{c|cccc} -2 & 1 & 0 & 6 & 20 \\ & \downa... & 1 & -2 & 10 & 0 \\ \end{array}$$ This gives \begin{align} P(z)&=(z+2)(z^2-2z+10)\\ &=(z+2)\left(z^2... near factors. $$P(z)=3z^2+7.$$ ====Solution==== \begin{align} P(z)&=3z^2+7\\ &=\left(\sqrt{3}z\right)^2-
- Question 8, Exercise 1.2
- ==== Assume $z=a+ib$, then $\overline{z}=a-ib$. \begin{align}z+\overline{z}&=\left( a+ib \right)+\left( ... Assume that $z=a+ib$, then $\overline{z}=a-ib$. \begin{align}z-\overline{z}&=\left( a+ib \right)-\left( ... Suppose $z=a+ib$, then $\overline{z}=a-ib$. Then \begin{align}z\overline{z}&=\left( a+ib \right)\cdot \le... e{z}&={{a}^{2}}+b^2. \ldots (1) \end{align} Now \begin{align} {{\left[\operatorname{Re}\left( z \right)
- Question 7, Exercise 1.1
- === We 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}\... === We know that $z_1=1+2i$ and $z_2=2+3i$, then \begin{align} {{z}_{1}}{{z}_{2}}&=\left( 1+2i \right)\ti... )+\left( 3+4 \right)i\\ &=-4+7i. \end{align} Now \begin{align} |z_1 z_2|&=\sqrt{(-4)^2+7^2}\\ &=\sqrt{16+
- Question 7, Exercise 1.2
- ry parts $\dfrac{2+3i}{5-2i}$. ====Solution==== \begin{align}&\dfrac{2+3i}{5-2i} \\ =&\dfrac{2+3i}{5-2i}... t( 1+2i \right)}^{2}}}{1-3i}$. ====Solution==== \begin{align}&\dfrac{(1+2i)^2}{1-3i}\\ =&\dfrac{1-4+4i}{... {{{\left( 1+i \right)}^{2}}}$. ====Solution==== \begin{align}&\dfrac{1-i}{{{\left( 1+i \right)}^{2}}}\\ ... {\left( 2a-bi \right)}^{-2}}$. ====Solution==== \begin{align}&{{\left( 2a-bi \right)}^{-2}}\\ =&\dfrac{1
- Question 6, Exercise 1.3
- ====Solution==== Given: $$z^3=-8.$$ This gives \begin{align} & z^3+2^3=0\\ \implies &(z+2)\left(z^2-2z+... a, we have $a=1$, $b=-2$ and $c=4$ Thus, we have \begin{align} z&=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}\\ &=\... $$(a-b)^3=a^3-b^3-3ab(a-b).$$ Therefore, we have \begin{align} &z^3-1-3z(z-1)=-1\\ \implies &{{z}^{3}}-1-... la, we have $a=1$, $b=-3$ and $c=3$. So, we have \begin{align}z&=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\
- Question 1, Exercise 1.1
- fy ${{i}^{9}}+{{i}^{19}}$. GOOD ====Solution==== \begin{align}{{i}^{9}}+{{i}^{19}}&=i\cdot{{i}^{8}}+i\cdo... \left( -i \right)}^{23}}$. GOOD ====Solution==== \begin{align}{{\left( -i \right)}^{23}}&={{\left( -1 \ri... right)}^{\frac{-23}{2}}}$. GOOD ====Solution==== \begin{align}{{\left( -1 \right)}^{\frac{-23}{2}}}&={{\l... \right)}^{\frac{15}{2}}}$. GOOD ====Solution==== \begin{align}{{\left( -1 \right)}^{\frac{15}{2}}}&={{\le
- Question 2 & 3, Exercise 1.1
- {i}^{122}}+{{i}^{153}}=0$. GOOD ====Solution==== \begin{align}L.H.S.&={{i}^{107}}+{{i}^{112}}+{{i}^{122}}... \right),-2\left( 1-3i \right)$. ====Solution==== \begin{align}& 3\left( 1+2i \right)+-2\left( 1-3i \right... }i,\dfrac{1}{4}-\dfrac{1}{3}i$. ====Solution==== \begin{align}&\left( \dfrac{1}{2}-\dfrac{2}{3}i \right)+... ht),\left( 1,\sqrt{2} \right)$. ====Solution==== \begin{align}&\left( \sqrt{2},1 \right)+\left( 1,\sqrt{2
- Question 6, Exercise 1.1
- the answer in the form $a+ib$. ====Solution==== \begin{align}\dfrac{4+i}{3+5i}&=\dfrac{4+i}{3+5i}\times ... the answer in the form $a+ib$. ====Solution==== \begin{align}\dfrac{1}{-8+i}&=\dfrac{1}{-8+i}\times \dfr... the answer in the form $a+ib$. ====Solution==== \begin{align}\dfrac{1}{7-3i}&=\dfrac{1}{7-3i}\times \dfr... the answer in the form $a+ib$. ====Solution==== \begin{align}\dfrac{6+i}{i}&=\dfrac{6+i}{i}\times \dfrac
- Question 9 & 10, Exercise 1.1
- right)\left( 2-i \right)}$. ====Solution==== Let \begin{align}z&=\dfrac{\left( 3-2i \right)\left( 2+3i \r... 2+2+4i-i}\\ &=\dfrac{12+5i}{4+3i}\end{align} Now \begin{align}\bar{z}&=\dfrac{12-5i}{4-3i}\\ &=\dfrac{12-... \right)}^{25}} \right]}^{3}}$. ====Solution==== \begin{align}i^{18}+\left(\dfrac{1}{i}\right)^{25} &=i^{... i \quad \because \dfrac{1}{i}=-i \end{align} Now \begin{align} {{\left[ {{i}^{18}}+{{\left( \dfrac{1}{i}
- Question 1, Exercise 1.2
- addition, that is, $$z_1+z_2=z_2+z_1.$$ We take \begin{align}z_1+z_2&=(2+i)+(1-i)\\ &=3 \ldots (i) \end{align} Now \begin{align} z_2+z_1&=(1-i)+(2+i)\\ &=3 \ldots (ii)\en... er multiplication, that is, $$z_1 z_2=z_2 z_1.$$ \begin{align}z_1 z_2 &=(2+i)\cdot(1-i) \\ &=(2+1)+(1-2)i\\ &=3-i \ldots (1) \end{align} Also \begin{align}z_2 z_1 &=(1-i)\cdot (2+i)\\ &=(2+1)+(1-2)i
- Question 6, Exercise 1.2
- rt{a^2+b^2}|$ and $|z_2=\sqrt{c^2+d^2}|$.\\ Now \begin{align} L.H.S.&=|{{z}_{1}}{{z}_{2}}|\\ &=|(a+bi)(c... Method**\\ We know $|z|^2=z\bar{z}$, so we have \begin{align}|{{z}_{1}}{{z}_{2}}{{|}^{2}}&={{z}_{1}}{{z}... se $z=a+bi$, then $|z|=\sqrt{a^2+b^2}$. We take \begin{align}\left| \dfrac{1}{z} \right|&=\left| \dfrac{... z} \right|&=\dfrac{1}{|z|} … (1)\end{align} Now \begin{align}L.H.S.&=\left| \dfrac{{{z}_{1}}}{{{z}_{2}}}
- Question 5, Exercise 1.3
- e have\\ $a=1$, $b=1$ and $c=3$.\\ Thus we have \begin{align}z&=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\... we have\\ $a=1$, $b=-1$ and $c=-1$\\ So we have \begin{align}z&=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\... we have\\ $a=1$, $b=-2$ and $c=i$\\ So we have \begin{align}z&=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\... =0$\\ ====Solution==== Given: $${{z}^{2}}+4=0$$ \begin{align} \implies &z^2-(2i)^2=0\\ \implies &(z+2i)