Question

If $z=\frac{(\sqrt{3}+i{)}^3(3i+4{)}^2}{(8+6i{)}^2}$, then $|z|$ is equal to

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is C $2$

Uisng the identity $|a+ib|=\sqrt{a^2+b^2}$
Since $|\frac{z_1{z}_2}{z_3}|=\frac{|z_1||z_2|}{|z_3|}$

Therefore

$|z|=|\frac{(\sqrt{3}+i{)}^3(3i+4{)}^2}{(8+6i{)}^2}|$

$=\frac{|(\sqrt{3}+i{)}^3||(3i+4{)}^2|}{|(8+6i{)}^2|}$

$=\frac{|\sqrt{3}+i{|}^3|3i+4{|}^2}{|8+6i{|}^2}$

$=\frac{2^3\cdot 5^2}{{10}^2}=2$