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NDA 2023
Exam
NDA
Year
2023
Subject
Mathematics
Chapter
Complex Number

NDA 2023 · Mathematics · Complex Number

If z is a complex number such that (z-1)/(z+1) is purely imaginary,...

If zz is a complex number such that z−1z+1\frac{z-1}{z+1} is purely imaginary, then what is ∣z∣|z| equal to?

Options

  1. A12\frac{1}{2}
  2. B23\frac{2}{3}
  3. C11
  4. D22

Correct answer

C. 11

Step-by-step solution

Let z=x+iyz−1z+1=(x−1)+iy(x+1)+iy=[(x−1)+iy]×[(x+1)−iy][(x+1)+iy]×[(x+1)−iy]=(x2−1)+y2+i[y(x+1)−y(x−1)](x+1)2+y2=(x2+y2−1)(x+1)2+y2+i(2y)(x+1)2+y2∵z=x+i y\frac{z-1}{z+1}=\frac{(x-1)+i y}{(x+1)+i y}=\frac{[(x-1)+i y] \times[(x+1)-i y]}{[(x+1)+i y] \times[(x+1)-i y]}=\frac{\left(x^2-1\right)+y^2+i[y(x+1)-y(x-1)]}{(x+1)^2+y^2}=\frac{\left(x^2+y^2-1\right)}{(x+1)^2+y^2}+i \frac{(2 y)}{(x+1)^2+y^2}\because It is purely imaginary, Re⁡(z−1z+1)=0x2+y2−1(x+1)2+y2=0∴x2+y2−1=0x2+y2=1∣z∣2=1∣z∣=1\operatorname{Re}\left(\frac{z-1}{z+1}\right)=0\frac{x^2+y^2-1}{(x+1)^2+y^2}=0\therefore \quad x^2+y^2-1=0x^2+y^2=1|z|^2=1|z|=1
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