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- Question 1, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- z+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 \\ \mat... \ 2z+3w&=2 …(ii)\end{align} Multiply $2$ by (i), we get\\ \begin{align}2z+2w&=6i …(iii)\end{align} Subtract (iii) from (ii), we get\\ $$\begin{array}{ccc} 2z & +2w & =6i\\ \ma
- Question 2, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- e exactly: $\sin \dfrac{\pi }{12}$ ===Solution=== We rewrite $\dfrac{\pi }{12}$ as $\dfrac{\pi }{3}-\d... luate exactly:$\tan {{75}^{\circ }}$ ==Solution== We rewrite ${{75}^{\circ }}$ as ${{45}^{\circ }}+{{3... uate exactly:$\tan {{105}^{\circ }}$ ==Solution== We rewrite ${{105}^{{}^\circ }}$ as ${{60}^{{}^\circ... ate exactly:$\tan \dfrac{5\pi }{12}$ ==Solution== We rewrite $\dfrac{5\pi }{12}$ as $\dfrac{\left( 2+3
- Question 6, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- $\dfrac{\theta }{2}=\dfrac{{{30}^{\circ }}}{2}$, we can find $\cos {{15}^{\circ }}$by using half angl... frac{\theta }{2}=\dfrac{{{135}^{\circ }}}{2}$, so we can find $\tan {{67.5}^{\circ }}$by using half an... frac{\theta }{2}=\dfrac{{{225}^{\circ }}}{2}$, so we can find $sin{{112.5}^{\circ }}$by using half ang... dfrac{\theta }{2}=\dfrac{\dfrac{\pi }{4}}{2}$, so we can find $\cos \dfrac{\pi }{8}$by using half angl
- Question 2, Exercise 10.3 @fsc-part1-kpk:sol:unit10
- circ }}+\sin {{43}^{\circ }}.$$ ====Solution==== We have an identity: $$\sin \alpha +\sin \beta =2\si... right)\end{align} Since $cos(-\theta)=cos\theta$, we have $$\sin {{37}^{\circ }}+\sin {{43}^{\circ }}=... \circ }}-\cos {{82}^{\circ }}$. ====Solution==== We have an identity: $$\cos \alpha -\cos \beta =-2\s... .\\ &=-2\sin(59^\circ)\sin(-23^\circ) \end{align} We have $\sin(-\theta)=-\sin\theta$, therefore $$\co
- Question 1, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- == Given ${{z}_{1}}=2+i$, ${{z}_{2}}=1-i$. First, we prove commutative property under addition, that is, $$z_1+z_2=z_2+z_1.$$ We take \begin{align}z_1+z_2&=(2+i)+(1-i)\\ &=3 \ld... )\\ &=3 \ldots (ii)\end{align} From (i) and (ii), we get required result. Now, we prove commutative property under multiplication, that is, $$z_1 z_2=z_2
- Question 2, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- i$, ${{z}_{2}}=3-2i$ and ${{z}_{3}}=2-2i$. First, we prove associative property under addition, that i... \ &=4-3i \ldots (2) \end{align} From (1) and (2), we have the required result. Now, we prove associative property under multiplication, that is, $$z_1 (z... +10)+(2+10)i\\ &=8+12i \ldots (3)\end{align} Now, we take \begin{align} z_1 z_2 &=(-1+i)\cdot (3-2i)\\
- Question 8, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- }^{2}}. \ldots (2) \end{align} Using (1) and (2), we get $$z\overline{z}={{\left[\operatorname{Re}\le... ose $z=a+bi$ ... (1) Then $\overline{z}=a-bi$. We have given \begin{align}&z=\overline{z} \\ \impli... i=0 \\ \implies &b=0 \end{align} Using it in (1), we get $z=a+0i=a$, that is, $z$ is real. =====Q... 2a=0\\ \implies &a=0 \end{align} Using it in (1), we get $$z=0+bi=bi,$$ that is, $z$ is pure imaginary
- Question 5, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- {2}}+z+3=0$\\ According to the quadratic formula, we have\\ $a=1,\,\,\,b=1$ and $c=3$\\ Quadratic for... {2}}-z-1=0$\\ According to the quadratic formula, we have\\ $a=1,\quad b=-1$ and $c=-1$\\ Quadratic f... 2}}-2z+i=0$\\ According to the quadratic formula, we have\\ $a=1,\,\,\,b=-2$ and $c=i$\\ Quadratic fo... }^{2}}+4=0$\\ According to the quadratic formula, we have\\ $a=1,\,\,\,b=0$ and $c=4$\\ Quadratic form
- Question 3, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- sin v=\dfrac{4}{5},$ $0\le v\le \dfrac{\pi }{2}.$ We know $\cos u=\pm \sqrt{1-{{\sin }^{2}}u}$ As $u$... },\,\,\,\,\,0\le v\le \dfrac{\pi }{2}.\end{align} We know $\cos u=\pm \sqrt{1-{{\sin }^{2}}u}$ \\ As $... 5},\,\,\,\, 0\le v\le \dfrac{\pi }{2}.\end{align} We know $\cos u=\pm \sqrt{1-{{\sin }^{2}}u}$\\ As $... sin v=\dfrac{4}{5},$ $0\le v\le \dfrac{\pi }{2}.$ We know $\cos u=\pm \sqrt{1-{{\sin }^{2}}u}$ As $u$
- Question 13, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- \varphi =\dfrac{3}{5}.\end{align} Thus, from (1), we have \begin{align}&4\sin\theta +3\cos\theta \\ &=... varphi =\dfrac{8}{17}.\end{align} Thus, from (1), we have \begin{align}&15\sin\theta +8\cos\theta \\ &... \dfrac{-5}{\sqrt{29}}.\end{align} Thus, from (1), we have \begin{align}&2\sin\theta -5\cos\theta \\ &=... =\dfrac{1}{\sqrt{2}}.\end{align} Thus, from (1), we have \begin{align}&\sin\theta+\cos\theta \\ &=\sq
- Question 2, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- } $$\implies \cos\theta = -\dfrac{12}{13}$$ Thus, we have the following by using double angle identiti... } $$\implies \cos\theta = -\dfrac{12}{13}$$ Thus, we have the following by using double angle identiti... } $$\implies \cos\theta = -\dfrac{12}{13}$$ Thus, we have the following by using double angle identities: Thus, we have the following by using double angle identiti
- Question 3, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- 1-kpk-ex10-2-q3.png?nolink |Reference triangle}} We find: $\cos \theta =-\dfrac{3}{5}$. Thus, we have the following by using double angle identity: \beg... 1-kpk-ex10-2-q3.png?nolink |Reference triangle}} We find: $\cos \theta =-\dfrac{3}{5}$. Thus, we have the following by using half angle identities: \be
- Question 1, Exercise 10.3 @fsc-part1-kpk:sol:unit10
- or difference $2\sin 6x\sin x$. ====Solution==== We have an identity: $$-2\sin \alpha \sin \beta =\co... \circ }}\cos {{123}^{\circ }}$. ====Solution==== We have an identity: $$2\sin \alpha \cos \beta =\si... c{A+B}{2}\cos \dfrac{A-B}{2}.$$ ====Solution==== We have an identity: $$2\sin \alpha \cos \beta =\s... c{P+Q}{2}\cos \dfrac{P-Q}{2}.$$ ====Solution==== We have an identity: $$2\cos \alpha \cos \beta =\co
- Question 7, Exercise 1.1 @fsc-part1-kpk:sol:unit01
- valuate $|{{z}_{1}}+{{z}_{2}}|$. ====Solution==== We know that $z_1=1+2i$ and $z_2=2+3i$, then \begin{... evaluate $|{{z}_{1}}{{z}_{2}}|$. ====Solution==== We know that $z_1=1+2i$ and $z_2=2+3i$, then \begin{... $\left|\dfrac{z_1}{z_2}\right|$. ====Solution==== We know that $z_1=1+2i$ and $z_2=2+3i$, then \begin{
- Question 11, Exercise 1.1 @fsc-part1-kpk:sol:unit01
- )(-2+i)\\ &=-4+1+2i+2i\\ &=-3+4i \end{align} Now we take \begin{align} \dfrac{z_1 z_2}{\overline{{z_1... &=\dfrac{-2}{5}+\dfrac{11i}{5} \end{align} Hence, we have $${\rm Re}\left( \dfrac{{{z}_{1}}{{z}_{2}}}{... frac{1}{z_1\overline{z_1}}=\dfrac{1}{5}.$$ Hence, we have $${\rm Im}\left(\dfrac{1}{z_1\overline{z_1}}