Question

Let$z_1=\frac{(\sqrt{3}+i{)}^2.(1-\sqrt{3}i)}{1+i}$, $z_2=\frac{(1+\sqrt{3}i{)}^2.(\sqrt{3}-i)}{1-i}$

• A. $|z_1|=|z_2|$
• B. $amp{z}_1+amp{z}_2=0$
• C. $3|z_1|=|z_2|$
• D. $3amp{z}_1+amp{z}_2=0$

Answer 1

Answer: (A), (D)

The correct options are
A $|z_1|=|z_2|$
D $3amp{z}_1+amp{z}_2=0$

$(i+\sqrt{3}{)}^2$
$=2+2\sqrt{3}i$
$=2(1+\sqrt{3}i)$
Hence
$\frac{(\sqrt{3}+i{)}^2(1-\sqrt{3}i)}{1+i}$
$=2\left(\frac{(1+\sqrt{3}i)(1-\sqrt{3}i)}{1+i}\right)$
$=2\left(\frac{(1-i)\left(4\right)}{2}\right)$
$=4\sqrt{2}\frac{1-i}{\sqrt{2}}$
$=4\sqrt{2}\left[cos\right(-{45}^0)+isin(-{45}^0\left)\right]$ ...(i)
Similarly
$z_2=-2\left[\frac{(1-\sqrt{3}i)(\sqrt{3}-i)}{1-i}\right]$
$=-2\left[\frac{(\sqrt{3}-i-3i-\sqrt{3}(1+i)}{2}\right]$
$=\left[\right(4i\left)\right(1+i\left)\right]$
$=4[i-1]$
$=4\sqrt{2}\left[cos\right({135}^0)+isin({135}^0\left)\right]$
Hence
$|z_1|=|z_2|=4\sqrt{2}$
$Arg\left(z_1\right)=-{45}^0$ and $Arg\left(z_2\right)={135}^0$
Hence
$3arg\left(z_1\right)+arg\left(z_2\right)=0$