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 \(\frac{\sqrt{41}}{4}\)
- C \(\frac{5}{3}\)
- D \(\frac{5}{4}\)
Answer & Solution
Correct Answer
(D) \(\frac{5}{4}\)
Step-by-step Solution
Detailed explanation
\(\text {Let } z=a+i b \)
\( \therefore |z|=\sqrt{a^2+b^2} \)
\( \therefore |z|+z=2+i \)
\( \therefore \sqrt{a^2+b^2}+a+i b=2+i\)
Comparing both sides, we get \(\sqrt{a^2+b^2}+a=2\) and \(b=1\)
\(\therefore \sqrt{1+a^2}+a=2 \)
\( \therefore 1+a^2=(2-a)^2 \)
\( \therefore 1+a^2=4-4 a+a^2 \)
\( \therefore 4 a=3\)
\(\therefore a=\frac{3}{4} \)
\( \therefore |z|=\sqrt{a^2+b^2}=\frac{5}{4}\)
\( \therefore |z|=\sqrt{a^2+b^2} \)
\( \therefore |z|+z=2+i \)
\( \therefore \sqrt{a^2+b^2}+a+i b=2+i\)
Comparing both sides, we get \(\sqrt{a^2+b^2}+a=2\) and \(b=1\)
\(\therefore \sqrt{1+a^2}+a=2 \)
\( \therefore 1+a^2=(2-a)^2 \)
\( \therefore 1+a^2=4-4 a+a^2 \)
\( \therefore 4 a=3\)
\(\therefore a=\frac{3}{4} \)
\( \therefore |z|=\sqrt{a^2+b^2}=\frac{5}{4}\)
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