Mathematics - Complex Number Question with Solution | TestHub

Since, $|z|+z=3+i$ Let $z=a+ib,$ then $$|z|+z=3+i\Rightarrow \sqrt{a^2+b^2}+a+ib=3+i$$ Compare real and imaginary coefficients on both sides $$\begin{array}{l}b=1,\sqrt{a^2+b^2}+a=3\\ \sqrt{a^2+1}=3-a\\ a^2+1=a^2+9-6a\\ 6a=8\Rightarrow a=\frac{4}{3}\end{array}$$ Then, $$|z|=\sqrt{{\left(\frac{4}{3}\right)}^2+1}=\sqrt{\frac{16}{9}+1}=\frac{5}{3}$$