Let z be a complex number such that
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:

1. (1) $\frac{\sqrt{34}}{3}$

2. (2) $\frac{5}{3}$

3. (3) $\frac{\sqrt{41}}{4}$

4. (4) $\frac{5}{4}$

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,

$|z|=\sqrt{\left(\frac{4}{3}\right)^{2}+1}=\sqrt{\frac{16}{9}+1}=\frac{5}{3}$

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