Question:
Let $\mathrm{z}$ be a complex number such that $|\mathrm{z}|+\mathrm{z}=3+\mathrm{i}$ $($ where $\mathrm{i}=\sqrt{-1})$.
Then $|z|$ is equal to:
Correct Option: , 2
Solution:
Since, $|z|+z=3+i$
Let $z=a+i b$, then
$|z|+z=3+i \Rightarrow \sqrt{a^{2}+b^{2}}+a+i b=3+i$
Compare real and imaginary coefficients on both sides
$b=1, \sqrt{a^{2}+b^{2}}+a=3$
$\sqrt{a^{2}+1}=3-a$
$a^{2}+1=a^{2}+9-6 a$
$6 a=8$
$a=\frac{4}{3}$
Then,
$|z|=\sqrt{\left(\frac{4}{3}\right)^{2}+1}=\sqrt{\frac{16}{9}+1}=\frac{5}{3}$