Baher asked in Science & MathematicsMathematics · 7 years ago

# solve the equation (z + 1)^5 = z^5 where z is a complex number?

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• kb
Lv 7
7 years ago

Taking 5th roots of both sides, we have

z + 1 = z * e^(2πik/5) with k = 0,1,2,3,4 [noting that e^(2πi/5) is a primitive 5th root of 1].

Solving for z:

z = 1/(e^(2πik/5) - 1), where k = 0, 1, 2, 3, 4.

I hope this helps!

• Ian H
Lv 7
7 years ago

Let w = r + is, such that w^5 = 1, then

(z + 1)^5 = z^5 = (zw)^5

(w - 1)z = 1

z = 1/[(r - 1) + is] = (r - 1)/[(r - 1)^2 + s^2] - is/[(r - 1)^2 + s^2] ...(1)

Using r = cos(2π/5) = (√5 - 1)/4 so that r - 1 = (√5 - 5)/4, and

s = sin(2π/5) = √[10 + 2√5 ]/4 and also

(r - 1)^2 + s^2 = [15 - 5√5 + 5 + √5]/8 = (5 - √5)/2

R(z) = [(√5 - 5)/4]/[(5 - √5)/2] = -1/2

I(z) = [√[10 + 2√5 ]/4]/[(5 - √5)/2] = [(5 + √5)^(3/2)*√2]/40

One value of z is given by z = -0.5 - i*0.688190960236..

A conjugate value of z is given by z = -0.5 + i*0.688190960236..

(This occurs when we replace 2π/5 by 8π/5)

With 4π/5 a similar calculation yields the same R(z), and the I(z) becomes

(√5 - 5)√[10 - 2√5 ]/40

This value of z is given by z = -0.5 - i*0.162459848116..

A conjugate value of z is given by z = -0.5 + i*0.162459848116..

(This occurs when we replace 4π/5 by 6π/5)

You may wonder about a possible 5th value for z.

10π/5 = 2π of course, so z = 1/(1 - 1) with

z = plus or minus infinity being theoretically equivalent fifth solutions

After trudging through that you may prefer KB's neater solution

z = 1/(e^(2πik/5) - 1), where k = 0, 1, 2, 3, 4.

Regards - Ian