What are the solutions to the equation (2-i)z^4+1-2i=0 ?

The solutions to the equation (2-i)z^4+1-2i=0 are

Find the zeros of (2-i)z^4+1-2i=0

The zeros of (2-i)z^4+1-2i=0 are

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

z = cos (\frac{π - arctan ( \frac{3}{4} )}{4}) + i sin (\frac{π - arctan ( \frac{3}{4} )}{4}), z = cos (\frac{- arctan ( \frac{3}{4} ) + 3 π}{4}) + i sin (\frac{- arctan ( \frac{3}{4} ) + 3 π}{4}), z = cos (\frac{- arctan ( \frac{3}{4} ) + 5 π}{4}) + i sin (\frac{- arctan ( \frac{3}{4} ) + 5 π}{4}), z = cos (\frac{- arctan ( \frac{3}{4} ) + 7 π}{4}) + i sin (\frac{- arctan ( \frac{3}{4} ) + 7 π}{4})

Solution steps

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

Move

2 i

to the right side

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

Move

1

to the right side

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

Divide both sides by

2 - i

z^{4} = - \frac{4}{5} + i \frac{3}{5}

For

z^{n} = a

the solutions are

k = 0, 1, \dots, n - 1

For

n = 4, a = - \frac{4}{5} + i \frac{3}{5}

∣ a ∣ = 1

ar g (a) = - arctan (\frac{3}{4}) + π

Simplify

z = cos (\frac{π - arctan ( \frac{3}{4} )}{4}) + i sin (\frac{π - arctan ( \frac{3}{4} )}{4}), z = cos (\frac{- arctan ( \frac{3}{4} ) + 3 π}{4}) + i sin (\frac{- arctan ( \frac{3}{4} ) + 3 π}{4}), z = cos (\frac{- arctan ( \frac{3}{4} ) + 5 π}{4}) + i sin (\frac{- arctan ( \frac{3}{4} ) + 5 π}{4}), z = cos (\frac{- arctan ( \frac{3}{4} ) + 7 π}{4}) + i sin (\frac{- arctan ( \frac{3}{4} ) + 7 π}{4})

λ^{3} - 14 λ^{2} + 9 λ - 110 = 0

z^{3} + 4 z^{2} - 16 z - 64 = 0

x^{2} (x + 2) (x + 6) = 0

625 = 5 x^{3}

2 x^{3} - 19 x^{2} + 64 x - 60 = 0

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