Question

If $a+ib+c{{i}^{2}}+d{{i}^{3}}={{\left( x+iy \right)}^{2}}$ then $\sqrt{a-ib-c{{i}^{2}}-d{{i}^{3}}}$ is equal to
A. –x + iy
B. –x – iy
C. x – iy
D. x + iy

Answer 1

Hint: We will start by simplifying the expression to us by using ${{i}^{3}}=-i\ and\ {{i}^{2}}=-1$. Then we will take the conjugate on both side of the equation and again use the properties like ${{i}^{3}}=-i\ and\ {{i}^{2}}=-1$ to form the term $a-ib+c{{i}^{2}}-d{{i}^{3}}$ in left hand side and further solve the equation to get the answer.

Complete step-by-step answer:
Now, we have been given that,
$a+ib+c{{i}^{2}}+d{{i}^{3}}={{\left( x+iy \right)}^{2}}$
Now, we know that the properties of i (iota) is,
$\begin{aligned} & {{i}^{2}}=-1 \\\ & {{i}^{3}}=-i \\\ \end{aligned}$
So, using this in LHS we have
$a+ib-c-id={{\left( x+iy \right)}^{2}}$
Now, we will conjugate on both sides. We know the conjugate of a complex number $a+ib$ is $a-ib$.
Therefore, we have,
$\left( a-c \right)-i\left( b-d \right)={{\left( x-iy \right)}^{2}}$
Now, expanding the terms in LHS we have,
$a-c-ib+id={{\left( x-iy \right)}^{2}}$
Now, we know that the fact that,
$\begin{aligned} & {{i}^{2}}=-1 \\\ & {{i}^{3}}=-i \\\ \end{aligned}$
So, using this we have,
$\begin{aligned} & a-ib+c{{i}^{2}}-d{{i}^{3}}={{\left( x-iy \right)}^{2}} \\\ & \sqrt{a-ib+c{{i}^{2}}-d{{i}^{3}}}=x-iy \\\ \end{aligned}$
Hence, the correct option is (C).

Note: To solve this question we have used the properties of i (iota) that $i\times i=-1\ and\ {{i}^{3}}=-i$. Also, it is important to note that we have used the conjugate of a complex number for converting a complex number $x+iy\ to\ x-iy$. Also, the conjugate of a complex number ${{z}^{2}}=\overline{{{z}^{2}}}=\left( \overline{{{z}^{2}}} \right)$. So, we can insert the conjugate inside the bracket from outside.