Question 14 Marks
Answer
View full question & answer→$i =|Z|=2 \cdot \sqrt{2}$
x = 2, y = -2
$\begin{array}{l}\cos \theta=\frac{z}{r}=\frac{2}{2 \sqrt{2}}=\frac{1}{\sqrt{2}} \\ \sin \theta=\frac{y}{r}=\frac{-2}{2 \sqrt{2}}=\frac{-1}{\sqrt{2}}\end{array}$
$\operatorname{Arg}(Z)=\frac{-\pi}{4}$
ii. $\bar{z} \bar{z}=|z|^2=(2 \sqrt{2})^2=8$
iii. $|Z|=\sqrt{2^2+(-2)^2}$
$=\sqrt{8}=2 \sqrt{2}$
OR
Real part of 2 - 2i = 2
x = 2, y = -2
$\begin{array}{l}\cos \theta=\frac{z}{r}=\frac{2}{2 \sqrt{2}}=\frac{1}{\sqrt{2}} \\ \sin \theta=\frac{y}{r}=\frac{-2}{2 \sqrt{2}}=\frac{-1}{\sqrt{2}}\end{array}$
$\operatorname{Arg}(Z)=\frac{-\pi}{4}$
ii. $\bar{z} \bar{z}=|z|^2=(2 \sqrt{2})^2=8$
iii. $|Z|=\sqrt{2^2+(-2)^2}$
$=\sqrt{8}=2 \sqrt{2}$
OR
Real part of 2 - 2i = 2



