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- Question 9, Exercise 1.2
- = \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 a... {400}\\ &= \frac{36}{25} \end{align} So, the real part of \(\left(\dfrac{4+2i}{2+5i}\right)^{-2}\) is \(\dfrac{17}{100}\) and the imaginary part is \(\dfrac{36}{25}\). ====Question 9(v)==== Find real and imaginary pa... frac{-720}{1681} \end{align} Therefore, the real part of \(\left(\dfrac{5 - 4i}{5 + 4i}\right)^2\) is \(\dfrac{-1519}{1681}\) and the imaginary pa
- 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} ... ither real or pure imaginary iff $(\overline{z})^{2}=z^{2}$. **Solution.** For $z=x+iy$, first sup
- 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