Question

If a > 0 and \[z=\dfrac{{{\left( 1+i \right)}^{2}}}{a-i}\] , has magnitude $\sqrt{\dfrac{2}{5}}$then $\overline{z}$ is equal to:
(a). $-\dfrac{3}{5}-\dfrac{1}{5}i$
(b). $-\dfrac{1}{5}+\dfrac{3}{5}i$
(c). $-\dfrac{1}{5}-\dfrac{3}{5}i$
(d). $\dfrac{1}{5}-\dfrac{3}{5}i$

Answer 1

Hint: While solving this question we will simplify the given $z=\dfrac{{{\left( 1+i \right)}^{2}}}{a-i}$ and using the magnitude of z is given as $\sqrt{\dfrac{2}{5}}$ . This will give us the value of z and this will conclude as with the value of $\overline{z}$ as we know that $\bar{z}=x-iy$ where $z=x+iy$.

Complete step by step answer:
From the question we have that
$z=\dfrac{{{\left( 1+i \right)}^{2}}}{a-i}$
We will now simplify this using ${{i}^{2}}=-1$ and ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$
And after some simple modifications we will multiply and divide the whole equation with $\left( a+i \right)$.
$\begin{align}
  & z=\dfrac{1+{{i}^{2}}+2i}{a-i} \\
 & \Rightarrow z=\dfrac{1-1+2i}{a-i} \\
 & \Rightarrow z=\dfrac{2i}{a-i}\left( \dfrac{a+i}{a+i} \right) \\
 & \Rightarrow z=\dfrac{2i\left( a+i \right)}{{{a}^{2}}-{{i}^{2}}} \\
 & \Rightarrow z=\dfrac{2i\left( a+i \right)}{{{a}^{2}}+1} \\
 & \Rightarrow z=\dfrac{2ai+2{{i}^{2}}}{{{a}^{2}}+1} \\
 & \Rightarrow z=-\dfrac{2}{{{a}^{2}}+1}+i\dfrac{2a}{{{a}^{2}}+1} \\
\end{align}$
Here we have the value of
$z=-\dfrac{2}{{{a}^{2}}+1}+i\dfrac{2a}{{{a}^{2}}+1}$
From the question it is given that the magnitude of $z$ is equal to $\sqrt{\dfrac{2}{5}}$.
As we know from the concept that the magnitude of $z=x+iy$ will be $\sqrt{{{x}^{2}}+{{y}^{2}}}$.
$\begin{align}
  & z=-\dfrac{2}{{{a}^{2}}+1}+i\dfrac{2a}{{{a}^{2}}+i} \\
 & \left| z \right|=\sqrt{{{\left( -\dfrac{2}{{{a}^{2}}+1} \right)}^{2}}+{{\left( \dfrac{2a}{{{a}^{2}}+1} \right)}^{2}}} \\
 & \Rightarrow \left| z \right|=\sqrt{\dfrac{4+4{{a}^{2}}}{{{\left( {{a}^{2}}+1 \right)}^{2}}}} \\
 & \Rightarrow \left| z \right|=\dfrac{2\sqrt{1+{{a}^{2}}}}{{{a}^{2}}+1} \\
 & \Rightarrow \left| z \right|=\dfrac{2}{\sqrt{{{a}^{2}}+1}} \\
\end{align}$
Here we have the magnitude of $\left| z \right|=\dfrac{2}{\sqrt{{{a}^{2}}+1}}$
By comparing this with given value in question we will get the value of a,
$\begin{align}
  & \sqrt{\dfrac{2}{5}}=\dfrac{2}{\sqrt{{{a}^{2}}+1}} \\
 & \Rightarrow \dfrac{1}{\sqrt{5}}=\dfrac{\sqrt{2}}{\sqrt{{{a}^{2}}+1}} \\
 & \Rightarrow \sqrt{{{a}^{2}}+1}=\sqrt{10} \\
 & \Rightarrow a=\pm 3 \\
\end{align}$
Given in the question that $a > 0$, so the value of a=+3
By using this value of a z will be
$z=-\dfrac{1}{5}+i\dfrac{3}{5}$
 As we know from the concept that the value will be $\bar{z}=x-iy$ for the $z=x+iy$.
$\bar{z}=-\dfrac{1}{5}-i\dfrac{3}{5}$

So, the correct answer is “Option C”.

Note: While solving these types of questions we should take care while performing the simplifications. We know that $\bar{z}=x-iy$ for the $z=x+iy$ not $\bar{z}=-x+iy$ for the $z=x+iy$.
Here from the concept we know that the magnitude of $z=x+iy$ will be $\sqrt{{{x}^{2}}+{{y}^{2}}}$ not${{x}^{2}}+{{y}^{2}}$. If we commit a small mistake even we will conclude with a complete wrong answer.