What are the solutions to the equation 4z^4-5iz^2+1=0 ?

The solutions to the equation 4z^4-5iz^2+1=0 are

Find the zeros of 4z^4-5iz^2+1=0

The zeros of 4z^4-5iz^2+1=0 are

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4 z^{4} - 5 i z^{2} + 1 = 0

Solution steps

4 z^{4} - 5 i z^{2} + 1 = 0

Refine

(4 z^{4} + 1) - 5 i z^{2} = 0

Write in the standard form

a_{n} x^{n} + \dots + a_{1} x + a_{0} = 0

4 z^{4} - 5 i z^{2} + 1 = 0

Rewrite the equation with

u = z^{2}

and

u^{2} = z^{4}

4 u^{2} - 5 i u + 1 = 0

Solve

4 u^{2} - 5 i u + 1 = 0 : u = - \frac{- 5 + 41}{8} i, u = \frac{5 + 41}{8} i

u = - \frac{- 5 + 41}{8} i, u = \frac{5 + 41}{8} i

Substitute back

u = z^{2},

solve for

z

Solve

The solutions are

x^{3} - 13 x^{2} + 59 x - 87 = 0

y, 2 x^{3} = 2 y^{2} + 5

x^{4} - 13 x^{2} - 12 x = 0

x^{4} - 200 x^{3} - 100 x^{2} = 0

x^{4} + 24 x^{3} + 122 x^{2} - 52 x + 25 = 0

What are the solutions to the equation 4z^4-5iz^2+1=0 ?
The solutions to the equation 4z^4-5iz^2+1=0 are
Find the zeros of 4z^4-5iz^2+1=0
The zeros of 4z^4-5iz^2+1=0 are