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- Question 1, Exercise 1.3
- oefficient. \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\\ \[\begi... thop+\limits_{-}11 \\ \hline 0&-11w&=11i &-11\\ \end{array} \] \begin{align}-11w&=11i-11\\ \implies w
- Question 5, Exercise 1.2
- i$. 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\\ &=3-i \ldots (2)\end{align} From (1) and (2), we have $$\overline{z_1... lign}z_1 z_2 &=(2+3i)(2-3i)\\ &=2^2-(3i)^2\\ &=13\end{align} Now \begin{align}\overline{z_1 z_2}=13 \ld
- Question 6, Exercise 1.3
- z}^{2}}+\dfrac{1}{2} \right)}^{2}}&=-\dfrac{3}{4}\end{align}\\ Take square root on both sides.\\ \begin... \pm \dfrac{\sqrt{3}}{2}i \right)}^{\dfrac{1}{2}}}\end{align} =====Question 6(ii)===== Find the solutio... {{a}^{2}}-ab+{{b}^{2}} \right)\\ z+2&=0\\ z&=-2\end{align} Now\\ $\left( {{z}^{2}}-2z+4 \right)=0$\... z&=\dfrac{2\pm 2\sqrt{3}i}{2}\\ z&=1\pm \sqrt{3}i\end{align} Then\\ $$z=-2,1\pm \sqrt{3}i$$ =====Quest
- Question 2, Exercise 1.2
- &=\left( -1+i \right)+\left( 3-2i \right)\\ &=2-i\end{align} So \begin{align} \left( {{z}_{1}}+{{z}_{2}... i \right)+\left( 2-2i \right)\\ &=4-3i \ldots (1)\end{align} Now \begin{align} {{z}_{2}}+{{z}_{3}}&=\left( 3-2i \right)+\left( 2-2i \right)\\ &=5-4i\end{align} So \begin{align} {{z}_{1}}+\left({{z}_{2}}... \right)+\left( 5-4i \right)\\ &=4-3i \ldots (2) \end{align} From (1) and (2), we have the required res
- Question 2, Exercise 1.3
- {align} P(-2)&=(-2)^3+6(-2)+20\\ &=-8-12+20\\ &=0\end{align} Thus $z+2$ is a factor of ${{z}^{3}}+6z+2... & -2 & 4 & -20 \\ \hline & 1 & -2 & 10 & 0 \\ \end{array}$$ This gives \begin{align} P(z)&=(z+2)(z^2... t[(z-1)^2-(3i)^2\right]\\ &=(z+2)(z-1+3i)(z-1-3i)\end{align} =====Question 2(ii)===== Factorize the po... qrt{7}i \right)\left( \sqrt{3}z-\sqrt{7}i \right)\end{align} =====Question 2(iii)===== Factorize the
- Question 8, Exercise 1.2
- +\overline{z}&=2\operatorname{Re}\left( z \right)\end{align} =====Question 8(ii)===== Show that $z-\o... {z}&=2bi\\ z-\overline{z}&=2i\operatorname{Im}(z)\end{align} =====Question 8(iii)===== Show that $z\o... mplies z\overline{z}&={{a}^{2}}+b^2. \ldots (1) \end{align} Now \begin{align} {{\left[\operatorname{Re... ) \right]}^{2}}&={{a}^{2}}+{{b}^{2}}. \ldots (2) \end{align} Using (1) and (2), we get $$z\overline{z}
- Question 7, Exercise 1.1
- {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}}=... 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+49}\\ &=\sqrt{65} \end{align} =====Question 7(iii)===== If ${{z}_{1}}=
- Question 7, Exercise 1.2
- rac{4+19i}{29}\\ =&\dfrac{4}{29}+\dfrac{19}{29}i \end{align} Real part $=\dfrac{4}{29}$\\ Imaginary par... frac{-15-5i}{10}\\ =&\dfrac{-3}{2}-\dfrac{1}{2}i\end{align} Real part $=\dfrac{-3}{2}$ \\ Imaginary p... \dfrac{-2-2i}{4}\\ =&-\dfrac{1}{2}-\dfrac{1}{2}i\end{align} Real part $=-\dfrac{1}{2}$\\ Imaginary par... abi}{{{\left( 4{{a}^{2}}+{{b}^{2}} \right)}^{2}}}\end{align} Real part $=\dfrac{4{{a}^{2}}-{{b}^{2}}}{{
- Question 1, Exercise 1.1
- \ &=i\cdot1+i\cdot\left( -1 \right)\\ &=i-i\\ &=0\end{align} =====Question 1(ii)===== Simplify ${{\l... \right)}^{11}}\\ &=-i\cdot\left( -1 \right)\\ &=i\end{align} =====Question 1(iii)===== Simplify ${{\le... rac{i}{i}\\ &=\frac{i}{-\left( -1 \right)}\\ &=i\end{align} =====Question 1(iv)===== Simplify ${{\le... ight)}^{7}}\\ &=i{{\left( -1 \right)}^{7}}\\ &=-i\end{align} ====Go to ==== <text align="left"><btn typ
- Question 2 & 3, Exercise 1.1
- left( -1 \right)}^{76}}\\ &=-i+1-1+i\\ &=0=R.H.S.\end{align} =====Question 3(i)===== Add the complex... =\left( 3-2 \right)+\left( 6+6 \right)i\\ &=1+12i\end{align} =====Question 3(ii)===== Add the complex... left( \dfrac{-2-1}{3} \right)i\\ &=\dfrac{3}{4}-i\end{align} =====Question 3(iii)===== Add the compl... ft( \sqrt{2}+1 \right)+\left( 1+\sqrt{2} \right)i\end{align} ====Go to ==== <text align="left"><btn t
- Question 6, Exercise 1.1
- 4}-\dfrac{17}{34}i\\ &=\dfrac{1}{2}-\dfrac{1}{2}i\end{align} =====Question 6(ii)===== Perform the in... rac{-8-i}{64+1}\\ &=\dfrac{-8}{65}-\dfrac{1}{65}i\end{align} =====Question 6(iii)===== Perform the in... \dfrac{7+3i}{58}\\ &=\dfrac{7}{58}+\dfrac{3}{58}i\end{align} =====Question 6(iv)===== Perform the in... }^{2}}}{-{{i}^{2}}}\\ &=\dfrac{-6i+1}{1}\\ &=1-6i\end{align} <text align="left"><btn type="primary">[[
- Question 9 & 10, Exercise 1.1
- frac{6+6+9i-4i}{2+2+4i-i}\\ &=\dfrac{12+5i}{4+3i}\end{align} Now \begin{align}\bar{z}&=\dfrac{12-5i}{4-... ac{63+16i}{25}\\ &=\dfrac{63}{25}+\dfrac{16}{25}i\end{align} =====Question 10===== Evalute ${{\left... ac{1}{i}\\ &=-1-i \quad \because \dfrac{1}{i}=-i \end{align} Now \begin{align} {{\left[ {{i}^{18}}+{{\... -i+3i-3 \right)\\ &=-\left( 2i-2 \right)\\ &=2-2i\end{align} ====Go to==== <text align="left"><btn typ
- Question 1, Exercise 1.2
- gin{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)\end{align} From (i) and (ii), we get required result.... )\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\\ &=3-i \ldots (2)\end{align} From (1) and (2), we get required result.
- Question 6, Exercise 1.2
- 2}}+{{d}^{2}}}\\ &=|{{z}_{1}}||{{z}_{2}}|=R.H.S. \end{align} **Alternative Method**\\ We know $|z|^2=z... ,\,\,|{{z}_{1}}{{z}_{2}}|&=|{{z}_{1}}||{{z}_{2}}|\end{align} proved. =====Question 6(ii)===== Show th... \left|\dfrac{1}{z} \right|&=\dfrac{1}{|z|} … (1)\end{align} Now \begin{align}L.H.S.&=\left| \dfrac{{{... (1)}\\ &=\dfrac{|{{z}_{2}}|}{|{{z}_{2}}|}=R.H.S.\end{align} ==== Go To ==== <text align="left"><btn
- Question 5, Exercise 1.3
- sqrt{11}}{2}i,-\dfrac{1}{2}-\dfrac{\sqrt{11}}{2}i\end{align} =====Question 5(ii)===== Find the solutio... \\ z&=\dfrac{1+\sqrt{5}}{2},\dfrac{1-\sqrt{5}}{2}\end{align} =====Question 5(iii)===== Find the soluti... z&=1\pm \sqrt{1-i}\\ z&=1+\sqrt{1-i},1-\sqrt{1-i}\end{align} =====Question 5(iv)===== Find the solutio... 2}\\ z&=\dfrac{\pm 4i}{2}\\ z&=\pm 2i\\ z&=2i,-2i\end{align} ====Go To==== <text align="left"><btn typ