Square Roots of Complex Numbers

Algebra 2 Tutorial

Square Roots of Complex Numbers


Just like for real numbers there will be two square roots for a complex number.

Finding these roots involves solving a system of two equations in two unknowns

Let’s say we want the square root of 3 + 4i.
The square root will be of the form a + bi

Then (a+bi)^2 = a^2 + 2abi - b^2
The minus sign in front of b squared comes from squaring i. Please notice that:
(-a-bi)^2 is also equal to a^2 + 2abi - b^2
and so we have a second root

a^2  - b^2 = 3 ……..(the real part)
2ab = 4 ……………(the imaginary part)

Sample Problem

Find the square root of 3 + 4i


[eq 1] a^2  - b^2 = 3 ……..(the real part)
[eq 2] 2ab = 4 ……………(the imaginary part)

a = 2/b …………from the eq 2
a^2 - 4/a^2 =3 …….substituting into eq1
[eq 3]a^4 - 3a^2 -4 =0 multiplying by a^2 and collecting terms on left hand side

(a^2 - 4) (a^2+1)=a^4+a^2 -4a^2 -4=a^4 - 3a^2 -4

so by eq 3
(a^2 - 4) (a^2+1)=0

a is the real part, so we can throw out a=i and
we are left with
(a^2 - 4)=0 and so
a=2 or a=-2 (two roots).

We know that 2ab = 4 (eq2).
So if a=2 then b=1.


(2+i)(2+i) = 4+2i+2i-1 .... -1 comes from i^2
(2+i)(2+i) = 3+4i ………….combining like terms

notice the second root is (-2+-i)
(-2-i)(-2-i) = 4+2i+2i-1 .... -1 comes from i^2
(-2-i)(-2-i) = 3+4i ………….combining like terms

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