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- Question 9, Exercise 1.2
- n. ====Question 9(i)==== Find real and imaginary parts of $(2+4 i)^{-1}$. **Solution.** Suppose $z=2+4i$. \... OD ====Question 9(ii)==== Find real and imaginary parts of $(3-\sqrt{-4})^{-2}$. **Solution.** Suppose $z=3 - \sqrt{-4}=3-2i... = \frac{12}{169}. \end{align} Therefore, the real part is \(\dfrac{5}{169}\) and the imaginary part is \(\dfrac{12}{169}\). GOOD ====Question 9(iii)==== Find real and imaginary parts of $\left(\dfrac{7+2 i}{3-i}\right)^{-1}$. **Solution.** We use the
- Question 4, Exercise 1.1
- al number $x$ and $y$ in each of the following: $(2+3i)x+(1+3i)y+2=0$ **Solution.** \begin{align}&(2+3i)x+(1+3i)y+2=0\\ \implies &(2x+y+2)+(3x+3y)i=0.\end{align} Comparing real and imaginary parts \beg
- Question 3, Exercise 1.4
- 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}\right)\left(x_{2}^{2}+y_{2}^{2}\right)\left(x_{3}^{2}+y_{3}^{2}\right) \ldots\left(x_{n}^
- Question 3, Exercise 1.2
- ====== Question 3, Exercise 1.2 ====== Solutions of Question 3 of Exercise 1.2 of Unit 01: Complex Numbers. This is unit of Model Te... e $\overline{z}=z$. Since $z$ is real, imaginary part of $z$ is zero. i.e. $b=0$. Then \begin{align} ... ightarrow \quad & a+ib=a-ib\\ \Rightarrow \quad & 2ib=0\\ \Rightarrow \quad & b=0\quad \because \quad
- Question 6(x-xvii), Exercise 1.4
- en complex number in the algebraic form: $7 \sqrt{2}\left(\cos \dfrac{5 \pi}{4}+i \sin \dfrac{5 \pi}{... ght)$ ** Solution. ** //Do yourself as previous parts.// =====Question 6(xi)===== Write a given complex number in the algebraic form: $10 \sqrt{2}\left(\cos \dfrac{7 \pi}{4}+i \sin \dfrac{7 \pi}{... t)$ ** Solution. ** //Do yourself as previous parts.// =====Question 6(xii)===== Write a given comp
- Question 8, Exercise 1.4
- }}{500}(1 +i). \end{align} Hence mean position of particle is $\frac{\sqrt{2}}{500}(1 +i)$. =====Question 8(ii)===== Calculate the position of a particle from mean position when amplitude is $0.004 \... (\dfrac{\pi}{3})) \\ &= \dfrac{4}{1000}(\dfrac{1}{2} + i \dfrac{\sqrt{3}}{2}) \\ &= \dfrac{1}{250} (\dfrac{1}{2} + i \dfrac{\sqrt{3}}{2}) \\ &= \dfrac{1
- Question 8, Review Exercise
- Islamabad, Pakistan. =====Question 8===== When particle is at a position of $\sqrt{2}+i \sqrt{2}$nm from its mean position calculate its amplitud... rc}$. ** Solution. ** Here we have $$x= \sqrt{2} + i \sqrt{2}, \quad \theta=\dfrac{\pi}{4}.$$ We have to find $x_{\max}$. By using the formula \begi... lign} &x=x_{\max} e^{i\theta} \\ \implies & \sqrt{2} + i \sqrt{2}=x_{\max} e^{i\dfrac{\pi}{4}} \\ \im
- Question 1, Review Exercise
- se> ii. Every complex number has $\operatorname{part}(\mathrm{s})$. * (a) one * (b) two ... ar{z}$ </collapse> v. In complex plane imaginary part is drawn along * (a) $x$-axis * (b) $... ="true">(b): $y$-axis</collapse> vi. If $z_{1}=3+2 i$ and $z_{2}=5+6 i$ then * (a) $z_{1}>z_{2}$ * (b) $z_{1}<z_{2}$ * %%(c)%% $\over
- Question 5, Exercise 1.1
- complex number $z$ if $4z-3\bar{z}=\dfrac{1-18i}{2-i}$ **Solution.** Suppose $z=x+iy$, then $\bar{... o we have \begin{align}&4z-3\bar{z}=\dfrac{1-18i}{2-i}\\ \implies &4(x+iy)-3(x-iy)=\dfrac{1-18i}{2-i}\times \dfrac{2+i}{2+i}\\ \implies &4x+4iy-3x+3iy=\dfrac{(1-18i)(2+i)}{2^2-i^2} \end{align} \begin
- Question 9, Exercise 1.4
- lamabad, Pakistan. =====Question 9(i)===== When particle is at a position of $x=2+3 i$ from its mean position and $x_{\max }=1+4 i$... ch problems.</fc> =====Question 9(ii)===== When particle is at a position of $x=2+3 i$ from its mean position and $x_{\max }=1+4 i$... can be seen under microscope at this point. If $x=2+3 i$ and $x_{\max }=1+4 i$. Calculate the frequency when $\mathrm{t}=2$. ** Solution. ** <fc #ff0000>The contents, giv
- Question 5, Exercise 1.4
- \ \sin \alpha + \sin \beta + \sin \gamma &= 0 -- (2) \end{align} Suppose $a=e^{i\alpha}$, $b=e^{i\be... eta + \sin \gamma). \\ \end{align} Using (1) and (2) above, we get $$a+b+c=0 -- (3).$$ Since we know \begin{align} &a^3+b^3+c^3-3abc \\ =&(a+b+c)(a^2+b^2+c^2-ab-bc-ca), \end{align} Using (3), we have $$a^3+b^3+c^3-3abc=0$$ $$\implies a^3+b^3+c^3=3abc