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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Complex Number

NDA 2024 · Mathematics · Complex Number

Let z_1 and z_2 be two complex numbers such that...

Let z1andz2z_1 and z_2 be two complex numbers such that \left|\frac{z_1+z_2}{z_1-z_2}\right|=1,thenwhatis, then what is\operatorname{Re}\left(\frac{z_1}{z_2}\right)+1$ equal to?

Options

  1. A-1
  2. B0
  3. C1
  4. D5

Correct answer

C. 1

Step-by-step solution

Let z1=x1+iy1andz2=x2+iy2z_1=x_1+i y_1 and z_2=x_2+i y_2 Now, ∣z1+z2z1−z2∣=1\quad\left|\frac{z_1+z_2}{z_1-z_2}\right|=1⇒∣z1+z2∣=∣z1−z2∣⇒∣(x1+x2)+i(y1+y2)∣=∣(x1−x2)+i(y1−y2)∣⇒(x1+x2)2+(y1+y2)2=(x1−x2)2+(y1−y2)2⇒x12+x22+2x1x2+y12+y22+2y1y2=x12+x22−2x1x2+y12+y22−2y1y2⇒x1x2+y1y2=0z1z2=x1+iy1x2+iy2×x2−iy2x2−iy2=(x1x2+y1y2)+i(y1x2−x1y2)x22+y22=0+i(y1x2−x1y2)x22+y22\begin{aligned} & \Rightarrow \quad\left|z_1+z_2\right|=\left|z_1-z_2\right| \\ & \Rightarrow\left|\left(x_1+x_2\right)+i\left(y_1+y_2\right)\right|=\mid\left(x_1-x_2\right) \\ & +i\left(y_1-y_2\right) \mid \\ & \Rightarrow\left(x_1+x_2\right)^2+\left(y_1+y_2\right)^2=\left(x_1-x_2\right)^2 \\ & +\left(y_1-y_2\right)^2 \\ & \Rightarrow x_1^2+x_2^2+2 x_1 x_2+y_1^2+y_2^2+2 y_1 y_2 \\ & =x_1^2+x_2^2-2 x_1 x_2+y_1^2 \\ & +y_2^2-2 y_1 y_2 \\ & \Rightarrow \quad x_1 x_2+y_1 y_2=0 \\ & \frac{z_1}{z_2}=\frac{x_1+i y_1}{x_2+i y_2} \times \frac{x_2-i y_2}{x_2-i y_2} \\ & =\frac{\left(x_1 x_2+y_1 y_2\right)+i\left(y_1 x_2-x_1 y_2\right)}{x_2^2+y_2^2} \\ & =0+\frac{i\left(y_1 x_2-x_1 y_2\right)}{x_2^2+y_2^2} \end{aligned}$\begin{aligned} & \therefore \quad \operatorname{Re}\left(\frac{z_1}{z_2}\right)=0 \\ & \therefore \quad \operatorname{Re}\left(\frac{z_1}{z_2}\right)+1=1.\end{aligned}
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