MHT CET · Maths · Complex Number
Let \(z\) be a complex number such that \(|z|+z=2+i\), where \(i=\sqrt{-1}\), then \(|z|\) is equal to
- A \(\frac{4}{5}\)
- B \(\frac{5}{4}\)
- C \(\frac{5}{3}\)
- D \(\frac{\sqrt{41}}{4}\)
Answer & Solution
Correct Answer
(B) \(\frac{5}{4}\)
Step-by-step Solution
Detailed explanation
\(|\mathrm{z}|+\mathrm{z}=2+\mathrm{i}\)
\(\Rightarrow \sqrt{x^2+y^2}+x+\mathrm{i} y=2+\mathrm{i}\)
Equating real and imaginary parts, we get
\(\sqrt{x^2+y^2}+x=2 \text { and } y=1 \)
\( \Rightarrow \sqrt{x^2+1}=2-x \)
\( \Rightarrow x^2+1=4-4 x+x^2 \)
\( \Rightarrow 4 x=3 \)
\( \Rightarrow x=\frac{3}{4} \)
\( \therefore \mathrm{z}=\frac{3}{4}+\mathrm{i} \)
\( \Rightarrow|z|=\sqrt{\left(\frac{3}{4}\right)^2+1^2}=\frac{5}{4}\)
\(\Rightarrow \sqrt{x^2+y^2}+x+\mathrm{i} y=2+\mathrm{i}\)
Equating real and imaginary parts, we get
\(\sqrt{x^2+y^2}+x=2 \text { and } y=1 \)
\( \Rightarrow \sqrt{x^2+1}=2-x \)
\( \Rightarrow x^2+1=4-4 x+x^2 \)
\( \Rightarrow 4 x=3 \)
\( \Rightarrow x=\frac{3}{4} \)
\( \therefore \mathrm{z}=\frac{3}{4}+\mathrm{i} \)
\( \Rightarrow|z|=\sqrt{\left(\frac{3}{4}\right)^2+1^2}=\frac{5}{4}\)
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