What are the solutions to the equation z^4=(1+2i)^8 ?

The solutions to the equation z^4=(1+2i)^8 are z=4+3i,z=-4-3i,z=-3+4i,z=3-4i

Find the zeros of z^4=(1+2i)^8

The zeros of z^4=(1+2i)^8 are z=4+3i,z=-4-3i,z=-3+4i,z=3-4i

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

z = 4 + 3 i, z = - 4 - 3 i, z = - 3 + 4 i, z = 3 - 4 i

Solution steps

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

Expand

(1 + 2 i)^{8} : - 527 + 336 i

z^{4} = - 527 + 336 i

Subtract

- 527 + 336 i

from both sides

z^{4} - (- 527 + 336 i) = - 527 + 336 i - (- 527 + 336 i)

Simplify

z^{4} + 527 - 336 i = 0

Rewrite the equation with

u = z^{2}

and

u^{2} = z^{4}

u^{2} + 527 - 336 i = 0

Solve

u^{2} + 527 - 336 i = 0 : u = 7 + 24 i, u = - 7 - 24 i

u = 7 + 24 i, u = - 7 - 24 i

Substitute back

u = z^{2},

solve for

z

Solve

z^{2} = 7 + 24 i : z = 4 + 3 i, z = - 4 - 3 i

Solve

z^{2} = - 7 - 24 i : z = - 3 + 4 i, z = 3 - 4 i

The solutions are

z = 4 + 3 i, z = - 4 - 3 i, z = - 3 + 4 i, z = 3 - 4 i

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

x^{4} + 4 x^{2} - 4 x^{3} - 4 x + 3 = 0

x^{5} = 2

x, 3 x^{3} + 2 y = 10

x^{3} - 15 x^{2} = 0

What are the solutions to the equation z^4=(1+2i)^8 ?
The solutions to the equation z^4=(1+2i)^8 are z=4+3i,z=-4-3i,z=-3+4i,z=3-4i
Find the zeros of z^4=(1+2i)^8
The zeros of z^4=(1+2i)^8 are z=4+3i,z=-4-3i,z=-3+4i,z=3-4i