Question

If $z_1$ and $z_2$ be two complex numbers then $\mathrm{Re}\left(z_1{z}_2\right)$ is equal to

• A. $\mathrm{Re}\left(z_1\right)\mathrm{Re}\left(z_2\right)$
• B. $\mathrm{Re}\left(z_1\right)\mathrm{Im}\left(z_2\right)$
• C. $\mathrm{Im}\left(z_1\right)\mathrm{Re}\left(z_2\right)$
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for the correct option.

Find the real part of $z_1{z}_2$.

Let $z_1=x+iy$ and $z_2=a+ib$ be two complex number.

Now, the real part of $z_1$ is $\mathrm{Re}\left(z_1\right)=x$ and its imaginary part is $\mathrm{Im}\left(z_1\right)=y$.

Again, the real part of $z_2$ is $\mathrm{Re}\left(z_2\right)=a$ and its imaginary part is $\mathrm{Im}\left(z_2\right)=b$.

The complex number $z_1{z}_2$ is given as:

$\begin{array}{rcl}{z}_1{z}_2& =& \left(x+iy\right)\left(a+ib\right)\\ & =& xa+ixb+iya+i^{2}yb\\ & =& xa+i\left(xb+ya\right)-yb\left[i^2=-1\right]\\ & =& \left(xa-yb\right)+i\left(xb+ya\right)\end{array}$

So the real part of $z_1{z}_2$ is $\mathrm{Re}\left(z_1{z}_2\right)=xa-yb$.

And it can be written in terms of real and imaginary parts of $z_1$ and $z_2$ as:

$\begin{array}{rcl}\mathrm{Re}\left(z_1{z}_2\right)& =& xa-yb\\ & =& \mathrm{Re}\left(z_1\right)\mathrm{Re}\left(z_2\right)-\mathrm{Im}\left(z_1\right)\mathrm{Im}\left(z_2\right)\end{array}$

So the real part of $z_1{z}_2$ is $\mathrm{Re}\left(z_1\right)\mathrm{Re}\left(z_2\right)-\mathrm{Im}\left(z_1\right)\mathrm{Im}\left(z_2\right)$.

Hence, the correct option is D.