Question

If $z=(1+i)(1-i\sqrt{3})(-2-2i)\left(i\right)\left(3\right),$ then

• A. $|z|=8$
• B. $\text{arg}\left(z\right)=-\frac{\pi }{3}$
• C. $|z|=24$
• D. $\text{arg}\left(z\right)=\frac{\pi }{6}$

Answer 1

Answer: (B), (C)

$|z|=24$

Given,
$z=(1+i)(1-i\sqrt{3})(-2-2i)\left(i\right)\left(3\right)$

$\text{arg}(1+i)=\frac{\pi }{4}$
$\text{arg}(1-i\sqrt{3})=-\frac{\pi }{3}$
$\text{arg}(-2-2i)=-\frac{3\pi }{4}$
$\text{arg}\left(i\right)=\frac{\pi }{2}$
$\text{arg}\left(3\right)=0$

$\therefore \text{arg}\left(z\right)=\text{arg}(1+i)+\text{arg}(1-i\sqrt{3})+\text{arg}(-2-2i)+\text{arg}\left(i\right)+\text{arg}\left(3\right)=\frac{\pi }{4}-\frac{\pi }{3}-\frac{3\pi }{4}+\frac{\pi }{2}+0=-\frac{\pi }{3}$

Now,
$|z|=|1+i|\cdot |1-i\sqrt{3}|\cdot |-2-2i|\cdot |i|\cdot |3|=\sqrt{1+1}\cdot \sqrt{1+3}\cdot \sqrt{4+4}\cdot \sqrt{1}\cdot \sqrt{9}=\sqrt{2}\cdot 2\cdot 2\sqrt{2}\cdot 1\cdot 3=24$