Question

If ${{z}_{1}}$, ${{z}_{2}}$ are $1-i$, $-2+4i$, respectively, find $\operatorname{Im}\left( \dfrac{{{z}_{1}}{{z}_{2}}}{\overline{{{z}_{1}}}} \right)$.

Answer 1

Hint: We will first multiply the given two complex numbers. We will find the conjugate of ${{z}_{1}}$. Then we will get a fraction $\dfrac{{{z}_{1}}{{z}_{2}}}{\overline{{{z}_{1}}}}$. We will rationalize this fraction by multiplying the numerator and denominator by the conjugate of the denominator. Then we will obtain a complex number of the form $\operatorname{Re}\left( \dfrac{{{z}_{1}}{{z}_{2}}}{\overline{{{z}_{1}}}} \right)+i\operatorname{Im}\left( \dfrac{{{z}_{1}}{{z}_{2}}}{\overline{{{z}_{1}}}} \right)$. From this we will get the desired answer.

Complete step by step answer:
We have ${{z}_{1}}=1-i$ and ${{z}_{2}}=-2+4i$. Now, if we have two complex numbers ${{x}_{1}}+i{{y}_{1}}$ and ${{x}_{2}}+i{{y}_{2}}$, then the multiplication of these two numbers is given as

$\begin{align}
  & \left( {{x}_{1}}+i{{y}_{1}} \right)\left( {{x}_{2}}+i{{y}_{2}} \right)={{x}_{1}}{{x}_{2}}+i{{y}_{1}}{{x}_{2}}+i{{y}_{2}}{{x}_{1}}+{{i}^{2}}{{y}_{1}}{{y}_{2}} \\
 & \Rightarrow \left( {{x}_{1}}+i{{y}_{1}} \right)\left( {{x}_{2}}+i{{y}_{2}} \right)={{x}_{1}}{{x}_{2}}+i{{y}_{1}}{{x}_{2}}+i{{y}_{2}}{{x}_{1}}-{{y}_{1}}{{y}_{2}} \\
 & \therefore \left( {{x}_{1}}+i{{y}_{1}} \right)\left( {{x}_{2}}+i{{y}_{2}} \right)=\left( {{x}_{1}}{{x}_{2}}-{{y}_{1}}{{y}_{2}} \right)+i\left( {{x}_{1}}{{y}_{2}}+{{x}_{2}}{{y}_{1}} \right) \\
\end{align}$
So, we get the following
$\begin{align}
  & {{z}_{1}}{{z}_{2}}=\left( 1-i \right)\left( -2+4i \right) \\
 & \Rightarrow {{z}_{1}}{{z}_{2}}=\left( 1\cdot \left( -2 \right)-\left( -1 \right)\cdot 4 \right)+i\left( 1\cdot 4+\left( -2 \right)\cdot \left( -1 \right) \right) \\
 & \Rightarrow {{z}_{1}}{{z}_{2}}=\left( -2+4 \right)+i\left( 4+2 \right) \\
 & \Rightarrow {{z}_{1}}{{z}_{2}}=\left( 2 \right)+i\left( 6 \right) \\
 & \therefore {{z}_{1}}{{z}_{2}}=2+6i \\
\end{align}$
Now, in the denominator we have $\overline{{{z}_{1}}}$, which is the conjugate of ${{z}_{1}}$. Therefore, we have ${{z}_{1}}=1+i$. So, we get the fraction as
$\dfrac{{{z}_{1}}{{z}_{2}}}{\overline{{{z}_{1}}}}=\dfrac{2+6i}{1+i}$
Now, we have to rationalize the fraction obtained. To do this, we will multiply the numerator and denominator by the conjugate of the denominator. The denominator is $\overline{{{z}_{1}}}$, so its conjugate is the complex number ${{z}_{1}}$ itself. So, we will rationalize the obtained fraction in the following manner,
$\dfrac{{{z}_{1}}{{z}_{2}}}{\overline{{{z}_{1}}}}=\dfrac{2+6i}{1+i}\times \dfrac{1-i}{1-i}$
Again, for multiplication in the numerator, we will use the following expression, $\left( {{x}_{1}}+i{{y}_{1}} \right)\left( {{x}_{2}}+i{{y}_{2}} \right)=\left( {{x}_{1}}{{x}_{2}}-{{y}_{1}}{{y}_{2}} \right)+i\left( {{x}_{1}}{{y}_{2}}+{{x}_{2}}{{y}_{1}} \right)$. For the expression in the denominator, we can see that it can be written using the algebraic identity $\left( a+b \right)\left( a-b \right)={{a}^{2}}-{{b}^{2}}$. So, using these identities, we get the following
$\dfrac{{{z}_{1}}{{z}_{2}}}{\overline{{{z}_{1}}}}=\dfrac{\left( 2\cdot 1-6\cdot -1 \right)+i\left( 2\cdot -1+6\cdot 1 \right)}{{{1}^{2}}-{{i}^{2}}}$
Simplifying the above fraction, we get
$\begin{align}
  & \dfrac{{{z}_{1}}{{z}_{2}}}{\overline{{{z}_{1}}}}=\dfrac{\left( 2+6 \right)+i\left( -2+6 \right)}{1-\left( -1 \right)} \\
 & \Rightarrow \dfrac{{{z}_{1}}{{z}_{2}}}{\overline{{{z}_{1}}}}=\dfrac{\left( 8 \right)+i\left( 4 \right)}{1+1} \\
 & \Rightarrow \dfrac{{{z}_{1}}{{z}_{2}}}{\overline{{{z}_{1}}}}=\dfrac{8+4i}{2} \\
 & \therefore \dfrac{{{z}_{1}}{{z}_{2}}}{\overline{{{z}_{1}}}}=4+2i \\
\end{align}$
Comparing the obtained simplification of the fraction with $\operatorname{Re}\left( \dfrac{{{z}_{1}}{{z}_{2}}}{\overline{{{z}_{1}}}} \right)+i\operatorname{Im}\left( \dfrac{{{z}_{1}}{{z}_{2}}}{\overline{{{z}_{1}}}} \right)$, we get that $\operatorname{Im}\left( \dfrac{{{z}_{1}}{{z}_{2}}}{\overline{{{z}_{1}}}} \right)=2$.

Note: If we have a complex number and its conjugate, then the conjugate of the conjugate is the complex number itself. That means $\overline{\left( \overline{z} \right)}=z$. The rationalization of a fraction works because of the algebraic identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$. Using this identity, we are able to eliminate the $i$ from the denominator and achieve simplification.