What are the solutions to the equation (z+1)^4=z^4 ?

The solutions to the equation (z+1)^4=z^4 are z=-1/2 ,z=-1/2+i 1/2 ,z=-1/2-i 1/2

Find the zeros of (z+1)^4=z^4

The zeros of (z+1)^4=z^4 are z=-1/2 ,z=-1/2+i 1/2 ,z=-1/2-i 1/2

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(z + 1)^{4} = z^{4}

z = - \frac{1}{2}, z = - \frac{1}{2} + i \frac{1}{2}, z = - \frac{1}{2} - i \frac{1}{2}

Solution steps

(z + 1)^{4} = z^{4}

Expand

(z + 1)^{4} : z^{4} + 4 z^{3} + 6 z^{2} + 4 z + 1

z^{4} + 4 z^{3} + 6 z^{2} + 4 z + 1 = z^{4}

Move

z^{4}

to the left side

4 z^{3} + 6 z^{2} + 4 z + 1 = 0

Factor

4 z^{3} + 6 z^{2} + 4 z + 1 : (2 z + 1) (2 z^{2} + 2 z + 1)

(2 z + 1) (2 z^{2} + 2 z + 1) = 0

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

2 z + 1 = 0 or 2 z^{2} + 2 z + 1 = 0

Solve

2 z + 1 = 0 : z = - \frac{1}{2}

Solve

2 z^{2} + 2 z + 1 = 0 : z = - \frac{1}{2} + i \frac{1}{2}, z = - \frac{1}{2} - i \frac{1}{2}

The solutions are

z = - \frac{1}{2}, z = - \frac{1}{2} + i \frac{1}{2}, z = - \frac{1}{2} - i \frac{1}{2}

0.15 = (1 + x)^{3} - 1

x (x^{3} + i) = i (x - 1) + 3

x^{3} + 6 x = 0

16 m^{4} + 48 m^{2} + 36 = 0

x^{3} - x^{2} - 400 = 0

What are the solutions to the equation (z+1)^4=z^4 ?
The solutions to the equation (z+1)^4=z^4 are z=-1/2 ,z=-1/2+i 1/2 ,z=-1/2-i 1/2
Find the zeros of (z+1)^4=z^4
The zeros of (z+1)^4=z^4 are z=-1/2 ,z=-1/2+i 1/2 ,z=-1/2-i 1/2