Question

If k>0, $\left| z \right|=\left| w \right|=k$ and $\alpha =\dfrac{z-\bar{w}}{{{k}^{2}}+z\bar{w}}$ then Re($\alpha$) equals to
a) 0
b) $\dfrac{k}{2}$
c) $k$
d) None of those.

Answer 1

We know that any complex number z has two parts, one real part and one imaginary part. We can write $z=x+iy$ where Re(z)=x and Im(z)=iy. We know that $\bar{z}=x-iy$ ,we can calculate the value of x and y using both the equation, we will get $x=\dfrac{z+\bar{z}}{2}$ and $y=\dfrac{z-\bar{z}}{2}$ i.e. Re(z)= $\dfrac{z+\bar{z}}{2}$ and Im(z)= $\dfrac{z-\bar{z}}{2}$. Similarly, for any conjugate number a, Re(a)= $\dfrac{a+\bar{a}}{2}$ and Im(a)= $\dfrac{a-\bar{a}}{2}$.
So Re($\alpha$)=$\dfrac{\alpha +\bar{\alpha }}{2}$. We will find $\bar{\alpha }$ and put it in the equation. We know that $z\bar{z}={{\left| z \right|}^{2}}$ ,we c\will get the value of k2 using this formula.

Complete step-by-step answer:
We have, $\alpha =\dfrac{z-\bar{w}}{{{k}^{2}}+z\bar{w}}$
Let z be any complex number,
$\Rightarrow z=x+iy........(i)$
We know that,
$\Rightarrow \bar{z}=x-iy........(ii)$
We will add both equation(i) and equation(ii), we get,
$\Rightarrow x=\dfrac{z+\bar{z}}{2}$
We can also say that $x=\dfrac{z+\bar{z}}{2}$ is nothing but real part of z ,so,
$\operatorname{Re}(z)=\dfrac{z+\bar{z}}{2}$
Similarly, we can say that real part of α,
$\operatorname{Re}(\alpha )=\dfrac{\alpha +\bar{\alpha }}{2}$
Multiplying equation(i) and equation(ii), we get,
$\begin{aligned} & \Rightarrow z\bar{z}=\left( x+iy \right)\left( x-iy \right) \\\ & \Rightarrow z\bar{z}={{x}^{2}}+{{y}^{2}} \\\ & \\\ \end{aligned}$
We know that ${{x}^{2}}+{{y}^{2}}={{\left| z \right|}^{2}}$,
$\Rightarrow z\bar{z}={{\left| z \right|}^{2}}......(iii)$
We have $\left| z \right|=\left| w \right|=k$,
We square the above equation, we will get,
$\Rightarrow {{\left| z \right|}^{2}}={{\left| w \right|}^{2}}={{k}^{2}}$
Using equation(iii), we can say that,
$\begin{aligned} & \Rightarrow {{\left| z \right|}^{2}}={{k}^{2}}=z\bar{z} \\\ & \Rightarrow {{\left| w \right|}^{2}}={{k}^{2}}=w\bar{w} \\\ \end{aligned}$
We know that $\operatorname{Re}(\alpha )=\dfrac{\alpha +\bar{\alpha }}{2}$ and we have , $\alpha =\dfrac{z-\bar{w}}{{{k}^{2}}+z\bar{w}}$ we will calculate $\bar{\alpha }$ and put it into equation,

& \Rightarrow \bar{\alpha }=\dfrac{\bar{z}-\bar{\bar{w}}}{{{k}^{2}}+\bar{z}\times \bar{\bar{w}}} \\\ & \\\ & \Rightarrow \bar{\alpha }=\dfrac{\bar{z}-w}{{{k}^{2}}+\bar{z}w} \\\ & \\\ & \Rightarrow \operatorname{Re}(\alpha )=\dfrac{1}{2}\left( \alpha +\bar{\alpha } \right) \\\ & \Rightarrow \operatorname{Re}(\alpha )=\dfrac{1}{2}\left( \dfrac{z-\bar{w}}{{{k}^{2}}+z\bar{w}}+\dfrac{\bar{z}-w}{{{k}^{2}}+\bar{z}w} \right) \\\ & \\\ & \Rightarrow \dfrac{\left( z-\bar{w} \right)\left( {{k}^{2}}+\bar{z}w \right)+\left( \bar{z}-w \right)\left( {{k}^{2}}+z\bar{w} \right)}{\left( {{k}^{2}}+z\bar{w} \right)\left( {{k}^{2}}+\bar{z}w \right)} \\\ \end{aligned}$$ $$\Rightarrow \dfrac{z{{k}^{2}}+z\bar{z}w-{{k}^{2}}\bar{w}-\bar{w}w\bar{z}+\bar{z}{{k}^{2}}+\bar{z}z\bar{w}-w{{k}^{2}}-zw\bar{w}}{\left( {{k}^{2}}+z\bar{w} \right)\left( {{k}^{2}}+\bar{z}w \right)}$$ We know that, $\begin{aligned} & \Rightarrow {{\left| z \right|}^{2}}={{k}^{2}}=z\bar{z} \\\ & \Rightarrow {{\left| w \right|}^{2}}={{k}^{2}}=w\bar{w} \\\ \end{aligned}$ We can solve the numerator using this, we get, $$\Rightarrow \dfrac{z{{k}^{2}}+{{k}^{2}}w-{{k}^{2}}\bar{w}-\bar{z}{{k}^{2}}+\bar{z}{{k}^{2}}+{{k}^{2}}\bar{w}-{{k}^{2}}w-z{{k}^{2}}}{\left( {{k}^{2}}+z\bar{w} \right)\left( {{k}^{2}}+\bar{z}w \right)}$$ We will get zero in the numerator so we can say that, $$\Rightarrow \operatorname{Re}(\alpha )=$$0 **So, the correct answer is “Option A”.** **Note:** The student must be clear with the concepts of complex numbers like conjugate of a complex number and modulus of a complex number. Don’t put the values of k, z, w in equation $\alpha =\dfrac{z-\bar{w}}{{{k}^{2}}+z\bar{w}}$ we have to calculate the real part of α not the value of α. Most of the time students get confused and make this mistake. Solve the expression carefully, try to eliminate as many variables as possible by substituting the values from the given condition.