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- Question 9, Exercise 1.2
- on 9(iii)==== Find real and imaginary parts of $\left(\dfrac{7+2 i}{3-i}\right)^{-1}$. **Solution.** We use the following formulas: \[Re\left(\left(\frac{x_1 + i y_1}{x_2 + i y_2}\right)^{-1}\right) = \frac{x_1 x_2 + y_1 y_2}{x_1^2 + y_1^2},\] \[Im\left(\left(\frac{x_1 + i y_1}{x_2 + i y_2}\right)^{-1}
- Question 7, Exercise 1.4
- x-1+iy) = -\dfrac{\pi}{4} \\ \implies & \tan^{-1}\left(\dfrac{y}{x-1}\right) = -\dfrac{\pi}{4} \\ \implies & \dfrac{y}{x-1} = \tan\left(-\dfrac{\pi}{4}\right) \\ \implies & \dfrac{y}{x-... s and inequations in Cartesian form: $z \bar{z}=4\left|e^{i \theta}\right|$ ** Solution. ** Suppose $z... en $\bar{z}=x-iy$. As \begin{align*} &z \bar{z}=4\left|e^{i \theta}\right| \\ \implies & (x+iy)(x-iy)=4|
- Question 10, Exercise 1.2
- Question 10(i)==== For $z_{1}=-3+2 i$, verify: $$\left|z_{1}\right|=\left|-z_{1}\right|=\left|\overline{z_{!}}\right|=\left|-\overline{z_{!}}\right|.$$ **Solution.** \begin{align} |z_1| &= \sqrt{
- Question 3, Exercise 1.4
- lamabad, Pakistan. =====Question 3(i)===== If $\left(x_{1}+i y_{1}\right)\left(x_{2}+i y_{2}\right)\left(x_{3}+i y_{3}\right) \ldots\left(x_{n}+i y_{n}\right)=a+i b$, show that: (i) $\left(x_{1}^{2}+y_{1}^{2
- Question 2, Exercise 1.4
- ====Question 2(i)===== Write the complex number $\left(\cos \dfrac{\pi}{6}+i \sin \dfrac{\pi}{6}\right)\left(\cos \dfrac{\pi}{3}+i \sin \dfrac{\pi}{3}\right)$... rac{\pi}{6}} \cdot e^{i\frac{\pi}{3}} \\ & = e^{i\left(\frac{\pi}{6}+\frac{\pi}{3}\right)} \\ & = e^{i\f... = 0 + i (1) = i. \end{align} Hence, we proved $$ \left(\cos \dfrac{\pi}{6}+i \sin \dfrac{\pi}{6}\right)\
- Question 1, Exercise 1.4
- end{align} and \begin{align} \alpha & = \tan^{-1}\left|\frac{y}{x}\right| = \tan^{-1}\left|\frac{2\sqrt{3}}{2}\right|\\ & = \tan^{-1}(\sqrt{3}) = \frac{\pi}{... \frac{\pi}{3}. \] Hence \[ 2 + i 2 \sqrt{3} = 4 \left( \cos \frac{\pi}{3} + i \sin \frac{\pi}{3} \right... {align} Next, \begin{align} \alpha &= \tan^{-1}\left|\frac{y}{x}\right| = \tan^{-1}\left|\frac{-\sqrt{
- Question 6(i-ix), Exercise 1.4
- n complex number in the algebraic form: $\sqrt{2}\left(\cos 315^{\circ}+i \sin 315^{\circ}\right)$ ** Solution. ** \begin{align} &\sqrt{2}\left(\cos 315^{\circ}+i \sin 315^{\circ}\right) \\ =& \sqrt{2} \left(\dfrac{1}{\sqrt{2}}-\dfrac{i}{\sqrt{2}} \right) \... a given complex number in the algebraic form: $5\left(\cos 210^{\circ}+i \sin 210^{\circ}\right)$ ** S
- Question 1, Exercise 1.1
- n{align}{{i}^{31}}&=i\cdot{{i}^{30}}\\ &=i\cdot{{\left( {{i}^{2}} \right)}^{15}}\\ &=i\cdot{{\left( -1 \right)}^{15}} \quad \because i^2=-1\\ &=i\cdot(-1)\\ &... ign} GOOD ====Question 1(ii)==== Evaulate ${{\left( -i \right)}^{6}}$. **Solution.** \begin{align} {{\left( -i \right)}^{23}}&=(-1)^{23} i^{23} \\ &=-1 \cdo
- Question 2, Exercise 1.2
- c properties of complex numbers to prove that $$ \left(z_{1} z_{2}\right)\left(z_{3} z_{4}\right)=\left(z_{1} z_{3}\right)\left(z_{2} z_{4}\right)=z_{3}\left(z_{1} z_{2}\right) z_{4} $$ **Solution.** \beg
- Question 6(x-xvii), Exercise 1.4
- complex number in the algebraic form: $7 \sqrt{2}\left(\cos \dfrac{5 \pi}{4}+i \sin \dfrac{5 \pi}{4}\rig... omplex number in the algebraic form: $10 \sqrt{2}\left(\cos \dfrac{7 \pi}{4}+i \sin \dfrac{7 \pi}{4}\rig... a given complex number in the algebraic form: $2\left(\cos\dfrac{5\pi}{2}+i \sin \dfrac{5\pi}{2}\right)... umber in the algebraic form: $\dfrac{1}{\sqrt{2}}\left(\cos \dfrac{\pi}{4}+i \sin \dfrac{\pi}{4}\right)$
- Question 8, Exercise 1.2
- . \end{align} Square both sides: \begin{align} & \left(5\sqrt{x^2 + (y + 4)^2}\right)^2 = \left(25 + 4y\right)^2 \\ \implies & 25 \left(x^2 + (y + 4)^2\right) = 625 + 200y + 16y^2. \\ \implies & 25 \left(x^2 + y^2 + 8y + 16\right) = 625 + 200y + 16y^2 \
- Question 4, Exercise 1.3
- ega$ in (1), we have \begin{align} &(1-i) z+(1+i)\left(\dfrac{2}{53}-\dfrac{7}{53}i \right)=3\\ \implies... on $(1)$ to find $z$: \begin{align} &2iz + (3-2i)\left( \dfrac{199}{205} + \dfrac{177}{205}i\right) = 1 ... {3}{i} z-(6+2 i) \omega=5 ; \quad \dfrac{i}{2} z+\left(\dfrac{3}{4}-\dfrac{1}{2} i\right) \omega=\left(\dfrac{1}{2}+2 i\right)$. ** Solution. ** Given: \begi
- Question 1, Exercise 1.3
- \begin{align} & z^2 + \dfrac{3}{25} \\ = & z^2 - \left(\dfrac{\sqrt{3}}{5} i \right)^2 \\ = & \left(z + \dfrac{\sqrt{3}}{5}i\right)\left(z - \dfrac{\sqrt{3}}{5}i\right) \end{align} ====Question 1(v)==== ... = 2 z^{3}+3 z^{2}-10 z-15.$$ Since \begin{align}P\left(-\dfrac{3}{2} \right) &= 2\left(-\dfrac{3}{2}\rig
- Question 6, Review Exercise
- bad, Pakistan. ===== Question 6===== Evaluate $\left[\dfrac{1}{i^{10}}+(2-i)^{2}+\sqrt{-25}\right]^{3}$ ** Solution. ** \begin{align*} &\left[\dfrac{1}{i^{10}} + (2 - i)^2 + \sqrt{-25}\right]^3\\ =&\left[\dfrac{1}{(i^2)^5} + ( 4 - 4i + i^2) + 5i \right]^3\\ =&\left[\dfrac{1}{(-1)^5} + ( 4 - 4i -1) + 5i \right]^3\\
- Question 1, Review Exercise
- {z}=3+4 i$ then $\mathrm{z}^{-1}$ is * (a) $\left(\frac{1}{3}, \frac{1}{4}\right)$ * (b) $\left(-\frac{1}{3},-\frac{1}{4}\right)$ * %%(c)%% $\left(\frac{3}{25}, \frac{-4}{25}\right)$ * (d) $\left(\frac{3}{25}, \frac{-4}{25}\right)$ \\ <btn type=