Question

Find the square roots of each complex number.
a. 5 + 12i
b. 5 − 12i

Answer 1

ANSWER:

Square root of 5 + 12i is 3+2i or -3 – 2i and square root of 5 – 12i is 3 – 2i or -3 + 2i.

SOLUTION:

Given, we have to find square roots of 5 + 12i and 5 – 12i

a) 5 + 12i

Now, square root of 5 + 12i  

Suppose that a + bi is a square root of 5 + 12i.  

$\text { Then, }(a+b i)^{2}=\left(a^{2}-b^{2}\right)+(2 a b) i=5+12 i$

Equate real and imaginary parts:

$\begin{array}{l}{a^{2}-b^{2}=5 \text { and } 2 a b=12 \rightarrow b=\frac{6}{a}} \\\\ {\text { So, } a^{2}-\left(\frac{6}{a}\right)^{2}=5} \\\\ {\rightarrow a^{2}-\frac{36}{a^{2}}=5} \\\\ {\rightarrow a^{4}-5 a^{2}-36=0} \\\\ {\rightarrow\left(a^{2}-9\right)\left(a^{2}+4\right)=0}\end{array}$

Since a must be real, a = 3 or -3.  

This gives b = 2 or -2, respectively.  

Thus, we have two square roots: 3+2i or -3-2i.

b) 5 - 12i

Now, square root of 5 - 12i  

Suppose that a - bi is a square root of 5 - 12i.  

$\text { Then, }(a-b i)^{2}=\left(a^{2}-b^{2}\right)-(2 a b) i=5-12 i$

Equate real and imaginary parts:

$\begin{array}{l}{a^{2}-b^{2}=5 \text { and } 2 a b=12 \rightarrow b=\frac{6}{a}} \\\\ {S o, a^{2}-\left(\frac{6}{a}\right)^{2}=5} \\\\ {\rightarrow a^{2}-\frac{36}{a^{2}}=5} \\\\ {\rightarrow a^{4}-5 a^{2}-36=0} \\\\ {\rightarrow\left(a^{2}-9\right)\left(a^{2}+4\right)=0} \\\\ {\text { since a must be real, } a=3 \text { or }-3}\end{array}$

This gives b = 2 or -2, respectively.  

Thus, we have two square roots: 3 - 2i or -3 + 2i.

Hence, square root of 5 + 12i is 3+2i or -3 – 2i and square root of 5 – 12i is 3 – 2i or -3 + 2i.