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- Question 4, Exercise 1.3
- (5+6i)\\ \implies &(2i+4) z + (6i-4) \omega = 10i-12 \\ \implies &(4+2i) z + (-4+6i) \omega = -12+10i \quad \cdots(4) \end{align} $(3)-(4)$,we have \begin{align} &(-1-8i+4-6i)\omega=3-i+12-10i\\ \implies &(3-14i)\omega=15-11i\\ \end{align... ts(2) \end{align} Multiply (1) by $2$ and (2) by $12$, we have \begin{align}-6i z - (12 + 4i) \omega =
- Question 9, Exercise 1.2
- -2 \cdot 3 \cdot (-2)}{(\sqrt{13})^4}\\ &= \frac{12}{169}. \end{align} Therefore, the real part is \(... frac{5}{169}\) and the imaginary part is \(\dfrac{12}{169}\). GOOD ====Question 9(iii)==== Find real a... x_1^2 - y_1^2\): \[x_1^2 - y_1^2 = 4^2 - 2^2 = 12.\] Now, compute \(4 x_2 x_1 y_2 y_1\): \[4 x_... t(x_1^2 + y_1^2\right)^2} \\ = &\frac{(-21) \cdot 12 + 320}{20^2}\\ = &\frac{-252 + 320}{400}\\ = &\fr
- Question 1, Exercise 1.4
- 2} = \sqrt{2^2 + (2\sqrt{3})^2} \\ & = \sqrt{4 + 12} = \sqrt{16} = 4. \end{align} and \begin{align} \... rt{3^2 + (-\sqrt{3})^2}\\ &= \sqrt{9 + 3} = \sqrt{12} = 2\sqrt{3}. \end{align} Next, \begin{align} \... ght) \right)\\ =&\sqrt{2} \left( \cos \frac{5\pi}{12}+ i \sin \frac{5\pi}{12} \right). \end{align} GOOD ====Go to ==== <text align="left"><btn typ
- Question 7, Exercise 1.1
- ==Question 7(i)==== Find the magnitude of the $11+12 i$. **Solution.** Suppose $$z=11+12i$$ Then \begin{align}|z|&= \sqrt{(11)^2+(12)^2}\\ &=\sqrt{265}\end{align} Hence $|11+12 i|=\sqrt{265}$. GOOD ====Question 7(ii)==== Find the m
- Question 2, Exercise 1.1
- $ **Solution.** \begin{align}&(3+2i)(4-3i)\\ =&12-9i+8i-6i^2\\ =&12+6-9i+8i\\ =&18-i\end{align} GOOD ====Question 2(vi)==== Write the following complex
- Question 9, Exercise 1.4
- =\dfrac{(2+3i)(1-4i)}{(1+4i)(1-4i)} \\ &=\dfrac{2+12-6i+3i}{1+16} \\ &=\dfrac{14}{17}-\dfrac{5}{17}i.