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- Question 8, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- t( z \right)$. ====Solution==== Assume $z=a+ib$, then $\overline{z}=a-ib$. \begin{align}z+\overline{z}&... \right)$. ====Solution==== Assume that $z=a+ib$, then $\overline{z}=a-ib$. \begin{align}z-\overline{z}&... ight]}^{2}}$. ====Solution==== Suppose $z=a+ib$, then $\overline{z}=a-ib$. Then \begin{align}z\overline{z}&=\left( a+ib \right)\cdot \left( a-ib \right)\\ &
- Question 9, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- Pakistan. =====Question 9(i)===== If $z=3+2i,$ then verify that $-|z|\leq \operatorname{Re}\left( z \... ight)\leq |z|$ ====Solution==== Given $z=3+2i$. Then $|z|=\sqrt{9+4}=\sqrt{13}$ and ${\rm Re}z=3=\sqrt... nd{align} =====Question 9(ii)===== If $z=3+2i,$ then verify that $-|z|\leq \operatorname{Im}\left( z \... ight)\leq |z|$ ====Solution==== Given $z=3+2i$. Then $|z|=\sqrt{9+4}=\sqrt{13}$ and ${\rm Im}z=2=\sqrt
- Question 7, Exercise 1.1 @fsc-part1-kpk:sol:unit01
- ution==== We know that $z_1=1+2i$ and $z_2=2+3i$, then \begin{align} {{z}_{1}}+{{z}_{2}}&=1+2i+2+3i\\ &=... ution==== We know that $z_1=1+2i$ and $z_2=2+3i$, then \begin{align} {{z}_{1}}{{z}_{2}}&=\left( 1+2i \ri... ution==== We know that $z_1=1+2i$ and $z_2=2+3i$, then \begin{align} \dfrac{z_1}{z_2}&=\dfrac{1+2i}{2+3i
- Question 5, Exercise 10.1 @fsc-part1-kpk:sol:unit10
- asure $\alpha$ nor $\beta$ in the first Quadrant, then find: $\sin \left( \alpha +\beta \right)$. ====... asure $\alpha$ nor $\beta$ in the first Quadrant, then find: $\cos \left( \alpha +\beta \right)$. ====... asure $\alpha$ nor $\beta$ in the first Quadrant, then find: $\tan \left( \alpha +\beta \right)$. ====
- Question 2, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- minal ray of $\theta $ is in the second quadrant, then find $\sin 2\theta $. ====Solution==== Given: $\... minal ray of $\theta $ is in the second quadrant, then find $\cos 2\theta $. ====Solution==== Given: $\... minal ray of $\theta $ is in the second quadrant, then find $\tan 2\theta $. ====Solution==== Given: $\
- Question 11, Exercise 1.1 @fsc-part1-kpk:sol:unit01
- ====Solution==== Given $z_1=2-i$ and $z_2=-2+i$, then $\overline{z_1}=2+i$. \begin{align} z_1 z_2&=(2-i... }}}} \right)$. ====Solution==== Given $z_1=2-i$, then $\overline{z_1}=2+i$. \begin{align} z_1\overline{
- Question 6, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- = Suppose ${{z}_{1}}=a+bi$ and ${{z}_{2}}=c+di$. Then $|z_1=\sqrt{a^2+b^2}|$ and $|z_2=\sqrt{c^2+d^2}|$... }_{2}}\ne 0$ ====Solution==== Suppose $z=a+bi$, then $|z|=\sqrt{a^2+b^2}$. We take \begin{align}\left
- Question 1, Review Exercise 1 @fsc-part1-kpk:sol:unit01
- e> v. If $z=x+iy$ and $|\dfrac{z-5i}{z+5i}|=1$, then $z$ lies on * (a) $X-axis$ * (b) $Y-a... If $\left( x+iy \right)\left( 2-3i \right)=4+i$, then * (a) $x=-\dfrac{14}{13},y=\dfrac{5}{13}$
- Question 3, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- rminal ray of $\theta$ is in the second quadrant, then find $\sin2\theta$. ====Solution==== Given: $\sin... rminal ray of $\theta$ is in the second quadrant, then find $\cos \dfrac{\theta }{2}$. ====Solution====
- Question 1, Review Exercise 10 @fsc-part1-kpk:sol:unit10
- llapse> ii. If$\tan {{15}^{\circ }}=2-\sqrt{3}$, then the value of ${{\cot }^{2}}{{75}^{\circ }}$ is ... t)=\dfrac{1}{2}$, and $\tan \alpha =\dfrac{1}{3}$ then $\tan \beta =$ * (a) $\dfrac{1}{6}$ *
- Question 1, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- ion 1===== If ${{z}_{1}}=2+i$and ${{z}_{2}}=1-i$, then verify commutative property w.r.t. addition and m
- Question 2, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- ==== $z_1=-1+i$, $z_2=3-2i$ and ${{z}_{3}}=2-2i$, then verify associative property w.r.t. addition and m
- Question 3 & 4, Exercise 1.2 @fsc-part1-kpk:sol:unit01
- {z}_{2}}=\sqrt{2}-\sqrt{3}i$and ${{z}_{3}}=2+3i$, then verify distributive property w.r.t. addition and
- Question 6, Exercise 1.3 @fsc-part1-kpk:sol:unit01
- \pm 2\sqrt{3}i}{2}\\ z&=1\pm \sqrt{3}i\end{align} Then\\ $$z=-2,1\pm \sqrt{3}i$$ =====Question 6(iii)==
- Question 4 and 5, Exercise 10.2 @fsc-part1-kpk:sol:unit10
- rminal ray of $\theta $ is in the third quadrant, then find $\sin \dfrac{\theta }{2}$. ====Solution====