=(1-2i)(1+2i)

Solution for =(1-2i)(1+2i) equation:

Simplifying
0 = (1 + -2i)(1 + 2i)

Multiply (1 + -2i) * (1 + 2i)
0 = (1(1 + 2i) + -2i * (1 + 2i))
0 = ((1 * 1 + 2i * 1) + -2i * (1 + 2i))
0 = ((1 + 2i) + -2i * (1 + 2i))
0 = (1 + 2i + (1 * -2i + 2i * -2i))
0 = (1 + 2i + (-2i + -4i2))

Combine like terms: 2i + -2i = 0
0 = (1 + 0 + -4i2)
0 = (1 + -4i2)

Solving
0 = 1 + -4i2

Solving for variable 'i'.

Move all terms containing i to the left, all other terms to the right.

Add '4i2' to each side of the equation.
0 + 4i2 = 1 + -4i2 + 4i2
Remove the zero:
4i2 = 1 + -4i2 + 4i2

Combine like terms: -4i2 + 4i2 = 0
4i2 = 1 + 0
4i2 = 1

Divide each side by '4'.
i2 = 0.25

Simplifying
i2 = 0.25

Take the square root of each side:
i = {-0.5, 0.5}