jmc

Question

All the complex roots of (z + 1)^5 = 32z^5, when plotted in the complex plane, lie on a circle. Find the radius of this circle.

Accepted Answer

Taking the absolute value of both sides, we get

∣ (z + 1)^{5} ∣ = ∣32 z^{5} ∣.

Then

∣ z + 1 ∣^{5} = 32∣ z ∣^{5},

so

∣ z + 1∣ = 2∣ z ∣.

Hence,

∣ z + 1 ∣^{2} = 4∣ z ∣^{2} .

Let

z = x + y i,

where

x

and

y

are real numbers. Then

∣ x + y i + 1 ∣^{2} = 4∣ x + y i ∣^{2},

which becomes

(x + 1)^{2} + y^{2} = 4 (x^{2} + y^{2}) .

This simplifies to

3 x^{2} - 2 x + 3 y^{2} + 1 = 0.

Completing the square, we get

(x - \frac{1}{3})^{2} + y^{2} = (\frac{2}{3})^{2} .

Thus, the radius of the circle is

\frac{2}{3} .