# What is (1 – 2i) ^6? Given the complex number ( 1-2i)^6

We need to simplify.

==> We will rewrite the power.

==> ( 1- 2i)^(2*3)

But we know that (x)^ab= (x^a)^b

==> (1-2i)^(2*3) = (1-2i)^2]^3

Now we will open the brackets.

==> (1 - 4i + 4i^2)^3

But i^2 =-1

==> ( 1- 4i -4)^3

==> ( -3 -4i)^3 = (-3-4i)^2 * (-3-4i)

= ( 9 +24i + 16i^2) * (-3-4i)

= ( 9+24i -16) * (-3-4i)

= ( -7+24i)*(-3-4i)

= ( 21 -72i +28i - 96i^2)

= ( 21+96 - 44i)

= ( 117 -44i)

==> ( 1-2i)^6 = 117 - 44*i

Approved by eNotes Editorial Team To find the value of (1 – 2i)^6 we use the DeMoivre's Theorem.

First convert (1 – 2i) to polar form, which is the form r [A]

r = sqrt (1^2 + -2^2) = sqrt (1 + 4) = sqrt 5

arc tan (2 / 1) = 63.43 degrees.

As (1 – 2i) is in the fourth quadrant we have A = 360 – 63.43 = 296.6 degrees.

So 1 – 2i = (sqrt 5) [A = 296.6]

(1 – 2i)^6

=> (sqrt 5) ^6 [A = 6*296.6]

=> 125 [A = 1779.39]

=> 125 [A = 339 degree]

Expressing this in the rectangular form:

x = 125 cos 339.39 = 117

y = 125 sin 339.39 = -44

Therefore (1 – 2i) ^6 = 117 – 44i

Approved by eNotes Editorial Team