Complex Numbers 2 Question 5
5. Let $z$ be a complex number such that $|z|+z=3+i$ (where $i=\sqrt{-1}$ ).
Then, $|z|$ is equal to
(2019 Main, 11 Jan II)
(a) $\frac{\sqrt{34}}{3}$
(b) $\frac{5}{3}$
(c) $\frac{\sqrt{41}}{4}$
(d) $\frac{5}{4}$
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Answer:
Correct Answer: 5. (b)
Solution:
- We have, $|z|+z=3+i$
$$ \begin{array}{ll} \text { Let } & z=x+i y \\ \therefore & \sqrt{x^{2}+y^{2}}+x+i y=3+i \\ \Rightarrow & \left(x+\sqrt{x^{2}+y^{2}}\right)+i y=3+i \\ \Rightarrow & x+\sqrt{x^{2}+y^{2}}=3 \text { and } y=1 \\ \text { Now, } & \sqrt{x^{2}+1}=3-x \\ \Rightarrow & x^{2}+1=9-6 x+x^{2} \\ \Rightarrow & 6 x=8 \Rightarrow x=\frac{4}{3} \\ \therefore & z=\frac{4}{3}+i \\ \Rightarrow & |z|=\sqrt{\frac{16}{9}+1}=\sqrt{\frac{25}{9}} \Rightarrow|z|=\frac{5}{3} \end{array} $$
