Question

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

Answer 1

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=(6)/(a)} \\\\ {\text { So, } a^(2)-\left((6)/(a)\right)^(2)=5} \\\\ {\rightarrow a^(2)-(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=(6)/(a)} \\\\ {S o, a^(2)-\left((6)/(a)\right)^(2)=5} \\\\ {\rightarrow a^(2)-(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.