Question

Find the square root of the complex number $-5+12i$?

Answer 1

We start solving the problem by assuming the square root of the given complex number. We then square them on both sides and compare the real and imaginary parts on both sides. We then make the necessary substitutions and calculations related to the real and imaginary parts of the square root to get the required result.

Complete step-by-step solution:
According to the problem, we need to find the square root of the complex number $-5+12i$.
Let us assume the square root of the given complex number $-5+12i$ be $a+bi$.
So, we get $a+bi=\sqrt{-5+12i}$ ---(1).
Let us do square on both sides.
$\Rightarrow {{\left( a+bi \right)}^{2}}=-5+12i$.
$\Rightarrow {{a}^{2}}+{{\left( ib \right)}^{2}}+2\left( a \right)\left( ib \right)=-5+12i$.
$\Rightarrow {{a}^{2}}+{{i}^{2}}{{b}^{2}}+2iab=-5+12i$.
We know that ${{i}^{2}}=-1$.
$\Rightarrow {{a}^{2}}+\left( -1 \right){{b}^{2}}+2iab=-5+12i$.
$\Rightarrow {{a}^{2}}-{{b}^{2}}+2iab=-5+12i$.
$\Rightarrow \left( {{a}^{2}}-{{b}^{2}} \right)+\left( 2ab \right)i=-5+12i$.
Comparing real and imaginary parts on both sides, we get ${{a}^{2}}-{{b}^{2}}=-5$ and $2ab=12$.
Now, we have $2ab=12\Leftrightarrow b=\dfrac{6}{a}$ ---(2). Let us substitute this result in ${{a}^{2}}-{{b}^{2}}=-5$.
So, we get ${{a}^{2}}-{{\left( \dfrac{6}{a} \right)}^{2}}=-5$.
$\Rightarrow {{a}^{2}}-\dfrac{36}{{{a}^{2}}}=-5$.
$\Rightarrow \dfrac{{{a}^{4}}-36}{{{a}^{2}}}=-5$.
$\Rightarrow {{a}^{4}}-36=-5{{a}^{2}}$.
$\Rightarrow {{a}^{4}}+5{{a}^{2}}-36=0$.
$\Rightarrow {{a}^{4}}+9{{a}^{2}}-4{{a}^{2}}-36=0$.
$\Rightarrow {{a}^{2}}\left( {{a}^{2}}+9 \right)-4\left( {{a}^{2}}+9 \right)=0$.
$\Rightarrow \left( {{a}^{2}}-4 \right)\left( {{a}^{2}}+9 \right)=0$.
$\Rightarrow {{a}^{2}}-4=0$ or ${{a}^{2}}+9=0$.
$\Rightarrow {{a}^{2}}=4$ or ${{a}^{2}}=-9$.
$\Rightarrow a=\pm 2$ (we know that ${{a}^{2}}$ is always greater than zero so, we neglect ${{a}^{2}}=-9$).
Now, let us find the value of b by substituting the value of “a” in equation (2).
So, we get $b=\dfrac{6}{\pm 2}\Leftrightarrow b=\pm 3$.
Let us substitute this in equation (1) to get the square roots of the given complex number.
So, we get $\pm \left( 2+3i \right)=\sqrt{-5+12i}$.
So, we have found the square root of the given complex number $-5+12i$ as $\pm \left( 2+3i \right)$.
$\therefore$ The square root of the complex number $-5+12i$ is $\pm \left( 2+3i \right)$.

Note: We should note that in a complex number $a+ib$, ‘a’ and ‘b’ are real numbers and this is the reason why we didn’t consider ${{a}^{2}}=-9$. We should not make calculation mistakes while solving this problem. We can also solve the problem by finding the exponential form of the given complex number and then find the square root of that exponential form which will require a good amount of calculation. Similarly, we can expect problems to find the Euler and exponential form of the given complex number.