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Let $z$ be a complex number such that $|z|+z=3+i$
$($ where $i=\sqrt{-1})$
Then $|\mathrm{z}|$ is equal to :
Options:
$($ where $i=\sqrt{-1})$
Then $|\mathrm{z}|$ is equal to :
Solution:
1261 Upvotes
Verified Answer
The correct answer is:
$\frac{5}{3}$
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,
$$
|z|=\sqrt{\left(\frac{4}{3}\right)^{2}+1}=\sqrt{\frac{16}{9}+1}=\frac{5}{3}
$$
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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