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

The solutions to the equation (z+1)^4=(z-1)^4 are z=0,z=i,z=-i

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

The zeros of (z+1)^4=(z-1)^4 are z=0,z=i,z=-i

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

z = 0, z = i, z = - i

Solution steps

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

Expand

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

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

Move

1

to the right side

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

Subtract

z^{4} - 4 z^{3} + 6 z^{2} - 4 z

from both sides

z^{4} + 4 z^{3} + 6 z^{2} + 4 z - (z^{4} - 4 z^{3} + 6 z^{2} - 4 z) = z^{4} - 4 z^{3} + 6 z^{2} - 4 z - (z^{4} - 4 z^{3} + 6 z^{2} - 4 z)

Simplify

8 z^{3} + 8 z = 0

Factor

8 z^{3} + 8 z : 8 z (z^{2} + 1)

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

Using the Zero Factor Principle:

ab = 0

then

a = 0

b = 0

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

Solve

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

The solutions are

z = 0, z = i, z = - i

y^{3} = 125

y^{3} = \frac{27}{1000}

x^{3} + x = 9

z^{4} = (1 + 2 i)^{8}

x, z = (2 x + 3 y)^{10}

What are the solutions to the equation (z+1)^4=(z-1)^4 ?
The solutions to the equation (z+1)^4=(z-1)^4 are z=0,z=i,z=-i
Find the zeros of (z+1)^4=(z-1)^4
The zeros of (z+1)^4=(z-1)^4 are z=0,z=i,z=-i