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JEE Main 201911 JanuaryEvening Shift

Exam: JEE Main
Year: 2019
Date: 11 January
Shift: Evening Shift
Subject: Mathematics
Chapter: Complex Number

JEE Main 2019 · 11 January · Evening Shift · Mathematics · Complex Number

Let z be a complex number such that |z|+z=3+i ( where i= -1 ) Then...

Let

z

be a complex number such that

|z|+z=3+i

(

where

i=\sqrt{-1})

Then

|\mathrm{z}|

is equal to :

Options

A $\frac{\sqrt{34}}{3}$
B $\frac{5}{3}$
C $\frac{\sqrt{41}}{4}$
D $\frac{5}{4}$

Correct answer

B. $\frac{5}{3}$

Step-by-step 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

\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}\right)^{2}+1}=\sqrt{\frac{16}{9}+1}=\frac{5}{3}

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