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- Question 4, Exercise 1.1
- 2+i)}=\dfrac{(1-5i)(2-i)+y(3-2i)}{(3-2i)(2-i)}\\ \Rightarrow \quad & \dfrac{x}{(2+i)}=\dfrac{(2+5i^2-10i-i)+(3y-2yi)}{(6+2i^2-4i-3i)}\\ \Rightarrow \quad & \dfrac{x}{(2+i)}=\dfrac{(2-5-11i)+(3y-2yi)}{(6+2i^2-4i-3i)}\\ \Rightarrow \quad & \dfrac{x}{(2+i)}=\dfrac{(3y-3)+(-11-2y)i}{(4-7i)}\\ \Rightarrow \quad & x(4-7i)=((3y-3)+(-11-2y)i)(2+i)\\ \Righta
- Question 3, Exercise 1.2
- n we $z$ is real. As \begin{align}& z=\bar{z}\\ \Rightarrow \quad & a+ib=a-ib\\ \Rightarrow \quad & 2ib=0\\ \Rightarrow \quad & b=0\quad \because \quad 2i\neq 0\end{align} Then (1) becomes $$z=a+i(0)=a
- Question 4, Exercise 1.2
- \end{align} Now \begin{align}&|z_{1} z_{2}|=16\\ \Rightarrow \quad &|z_{1}|| z_{2}|=16\\ \Rightarrow \quad & \sqrt{13}|z_2|=16\\ \Rightarrow \quad & |z_2|=\dfrac{16}{\sqrt{13}}\end{align} GOOD ====Go to ==== <