JEE Mains · Maths · STD 11 - 4.1 complex nubers
Let \(z\,\ne -i\) be any complex number such that \(\frac{{z - i}}{{z + i}}\) is a purely imaginary number. Then \(z +\frac {1}{z}\) is
- A \(0\)
- B any non-zero real number other than \(1.\)
- C any non-zero real number.
- D a purely imaginary number
Answer & Solution
Correct Answer
(C) any non-zero real number.
Step-by-step Solution
Detailed explanation
Let \(z=x+i y\) \(\frac{z-i}{z+i}\) is purely imaginary means its real part is zero. \(\frac{x+i y-i}{x+i y+i}=\frac{x+i(y-1)}{x+i(y+1)} \times \frac{x-i(y+1)}{x-i(y+1)}\) \({ = \frac{{{x^2} - 2ix(y + 1) + xi(y - 1) + {y^2} - 1}}{{{x^2} + {{(y + 1)}^2}}}}\)…
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