Question

Find the square root of $7-24i$.

Answer 1

Hint : We find the square root of the given complex number $7-24i$. The square root is considered as the value of the variable $x$. Then we use the conjugate theorem to find the other root for the equation. We use the quadratic equation ${{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}}$ to find the required equation.

Complete step by step solution:
We express $7-24i$ in the form of identity form of ${{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}}$.
For our given expression $7-24i$, we convert 7 for the form ${{a}^{2}}+{{b}^{2}}$ of ${{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}}$. Then we convert $24i$ for the form $2ab$ of ${{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}}$.
We break 7 as $7=16-9=16+9{{i}^{2}}={{4}^{2}}+{{\left( 3i \right)}^{2}}$. We have the sum of two squares.
$\begin{aligned} & 7-24i \\\ & ={{4}^{2}}+{{\left( 3i \right)}^{2}}-2\times 4\times 3i \\\ \end{aligned}$
For our identity ${{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}}$, we got $a=4,b=3i$.
So, $7-24i={{\left( 4-3i \right)}^{2}}$.
We can express the root as $\sqrt{7-24i}=\pm \left( 4-3i \right)$.
So, the correct answer is “$\pm \left( 4-3i \right)$”.

Note : The equation ${{x}^{2}}-\left( p+q \right)x+pq=0$ can be broken into two parts where $\left( x-p \right)\left( x-q \right)=0$ giving two roots as $p$ and $q$. In our given problem the value of $p$ and $q$ are both the same. The “i” is an imaginary number in x+iy.