Question

If z₁ = 1 - i  and z₂ = i then \(\rm Re \left(\frac{\bar {z_{1}}z_{2}}{z_{1}} \right)=?\)
1. 1
2. -1
3. 0
4. 2

Answer 1

Correct Answer - Option 2 : -1

Concept:

Let z = x + iy be a complex number,

Where x is called the real part of the complex number or Re (z) and y is called the Imaginary part of the complex number or Im (z)

Conjugate of z = z̅ = x - iy  

 

Calculation:

z₁ = 1 - i 

z₂ = i 

Conjugate of z₁ = 1 + i̅

\(\Rightarrow \frac{\overline z_1z_2}{z_1}=\frac{(1+i)i}{1-i}\)

\(=\frac{i+i^2}{1-i}\)

\(=\frac {-1+i}{1-i}\)

Multiplying numerator and denominator by  1 + i

\(=\frac{-1+i}{1-i}\times \frac{1+i}{1+i}\)

\(=\frac {(-1+i)(1+i)}{1-i^2}\)

\(=\frac{-1-i+i+i^2}{1+1}\)

\(=\frac {-2}{2}\)

\(=-1\)

So, \(\rm Re \left(\frac{\bar {z_{1}}z_{2}}{z_{1}} \right)=-1\)

Hence, option 2 is correct