Question

If $\sqrt{x+iy}=\pm (a+ib),$ then $\sqrt{-x-iy}$ is equal to
1) $\pm (b+ia)$
2) $\pm (a-ib)$
3) $\pm (b-ia)$
4) None of these

Answer 1

Hint: Here, first we will simplify the equation $\sqrt{x+iy}=\pm (a+ib),$ so that the square roots are removed. After, that we will equate the real parts and imaginary parts of the simplified complex numbers obtained. This result will be further used to solve the problem.

Complete Step-by-Step solution:
A complex number $z$ has two parts. A real part and an imaginary part. If $z=a+ib$ and $z=c+id$ are two complex numbers then, $a+ib=c+id.$ This implies,
$\begin{align}
  & \Rightarrow a=c \\
 & \Rightarrow b=d \\
\end{align}$
Here, $a,c$ and $b,d$ are the real and imaginary parts of the complex numbers $a+ib$ and $c+id$ respectively. These properties can be applied in this problem to get the correct answer.
The symbol $i$(iota) is used to represent the square root of $\sqrt{-1}$. This also implies, ${{i}^{2}}=-1$, which means $i$ is the solution of the quadratic equation ${{x}^{2}}+1=0.$ In this way square root of any negative number can be expressed using $i.$
$\begin{align}
  & \Rightarrow \sqrt{-1}=i.........(i) \\
 & \Rightarrow {{i}^{2}}=-1.......(ii) \\
\end{align}$
Here, in this question we have to find $\sqrt{-x-iy}$ given that $\sqrt{x+iy}=\pm (a+ib).$
It is given that $\sqrt{x+iy}=\pm (a+ib).$
$\Rightarrow \sqrt{(x+iy)}=\pm (a+ib)........(iii)$
Now, squaring equation (iii) on both sides we get,
$\begin{align}
  & \Rightarrow {{\left( \sqrt{(x+iy)} \right)}^{2}}=\pm {{\left( (a+ib) \right)}^{2}} \\
 & \Rightarrow x+iy=\pm \left( {{a}^{2}}+2\times a\times ib+{{(ib)}^{2}} \right) \\
\end{align}$
$\Rightarrow x+iy=\pm ({{a}^{2}}+i2ab+{{i}^{2}}{{b}^{2}}).......(iv)$
From equation (ii) we know that ${{i}^{2}}=-1$.
Hence, we can rewrite equation (iv) as,
\[\Rightarrow x+iy=\pm ({{a}^{2}}+i2ab-{{b}^{2}}).........(v)\]
Rearranging equation (v) we get,
\[\Rightarrow x+iy=\pm ({{a}^{2}}-{{b}^{2}}+i2ab).........(vi)\]
We also know that if $a+ib$ and $c+id$ are two complex numbers and if, $a+ib=c+id.$ This implies,
$\begin{align}
  & \Rightarrow a=c \\
 & \Rightarrow b=d \\
\end{align}$
Comparing this with equation (vi) we get,
$\begin{align}
  & \Rightarrow x=\pm ({{a}^{2}}-{{b}^{2}}).........(vii) \\
 & \Rightarrow y=\pm (2ab)..........(viii) \\
\end{align}$
In the given question we have to find $\sqrt{-x-iy}.$ Now, squaring this expression we get,
$\Rightarrow {{\left( \sqrt{-x-iy} \right)}^{2}}=-x-iy.........(ix)$
From equations (vii) and (viii) we know that,
$\begin{align}
  & \Rightarrow -x=\pm (-({{a}^{2}}-{{b}^{2}})) \\
 & \Rightarrow -x=\pm ({{b}^{2}}-{{a}^{2}})........(x) \\
 & \Rightarrow -y=\pm (-2ab)........(xi) \\
\end{align}$
Now substituting equations (x) and (xi) in equation (ix) we get,
$\Rightarrow -x-iy=\pm ({{b}^{2}}-{{a}^{2}}-i2ab).......(xii)$
From equation (iv) we know that,
$\Rightarrow \pm {{\left( a+ib \right)}^{2}}=\pm ({{a}^{2}}+2iab-{{b}^{2}})$
Comparing this with equation (xii) we can rewrite $\pm ({{b}^{2}}-{{a}^{2}}-i2ab)$ as,
$\begin{align}
  & \Rightarrow \pm {{(b-ia)}^{2}}=\pm ({{b}^{2}}-2iab+{{\left( ia \right)}^{2}}) \\
 & \Rightarrow {{(b-ia)}^{2}}=\pm ({{b}^{2}}-2iab-{{a}^{2}})=\pm ({{b}^{2}}-{{a}^{2}}-i2ab).....(xiii) \\
\end{align}$
$\begin{align}
  & \Rightarrow \sqrt{-x-iy}=\pm \sqrt{{{(b-ia)}^{2}}} \\
 & \Rightarrow \sqrt{-x-iy}=\pm (b-ia).......(xiv) \\
\end{align}$
$\therefore $ the correct option is (3).

Note: In this problem, the main property of complex number used is, if $a+ib$ and $c+id$ are two complex numbers and if, $a+ib=c+id.$
Then,
$\begin{align}
  & \Rightarrow a=c \\
 & \Rightarrow b=d \\
\end{align}$
There are chances that a student might write $\sqrt{-x-iy}$ as $\sqrt{-1(x+iy)}$ and then again rewrite it as $\sqrt{i(x+iy}.$ This is wrong. It should be noted that $\sqrt{-1}=i$ and hence you cannot write like this.