Let $z$ be a complex number such that $|z|+z=3+i$ $($ where $i=\sqrt{-1})$ Then $|\mathrm{z}|$ is equal to :

Question

Aicharya · Accepted Answer

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 $ \begin{array}{l} 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 \Rightarrow a=\frac{4}{3} \end{array} $ Then, $ |z|=\sqrt{\left(\frac{4}{3}ight)^{2}+1}=\sqrt{\frac{16}{9}+1}=\frac{5}{3} $

Let $z$ be a complex number such that $|z|+z=3+i$ $($ where $i=\sqrt{-1})$ Then $|\mathrm{z}|$ is equal to :

Solution