What are the solutions to the equation z^3=2^{5/2}(1+i) ?

The solutions to the equation z^3=2^{5/2}(1+i) are

Find the zeros of z^3=2^{5/2}(1+i)

The zeros of z^3=2^{5/2}(1+i) are

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z^{3} = 2^{\frac{5}{2}} (1 + i)

z = \frac{1 + 3 ^{\frac{1}{2}}}{2 ^{\frac{1}{2}}} + i \frac{- 1 + 3}{2 ^{\frac{1}{2}}}, z = - 2^{\frac{1}{2}} + 2^{\frac{1}{2}} i, z = \frac{1 - 3 ^{\frac{1}{2}}}{2 ^{\frac{1}{2}}} + i \frac{- 1 - 3}{2 ^{\frac{1}{2}}}

Solution steps

z^{3} = 2^{\frac{5}{2}} (1 + i)

Simplify

2^{\frac{5}{2}} (1 + i) : 4 \cdot 2^{\frac{1}{2}} + 4 \cdot 2^{\frac{1}{2}} i

For

z^{n} = a

the solutions are

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

For

n = 3, a = 4 \cdot 2^{\frac{1}{2}} + 4 \cdot 2^{\frac{1}{2}} i

∣ a ∣ = 8

ar g (a) = \frac{π}{4}

Simplify

z = \frac{1 + 3 ^{\frac{1}{2}}}{2 ^{\frac{1}{2}}} + i \frac{- 1 + 3}{2 ^{\frac{1}{2}}}, z = - 2^{\frac{1}{2}} + 2^{\frac{1}{2}} i, z = \frac{1 - 3 ^{\frac{1}{2}}}{2 ^{\frac{1}{2}}} + i \frac{- 1 - 3}{2 ^{\frac{1}{2}}}

4 x (x^{2} - 4) = 0

x, f = x^{3}

(1 - x) (1 - x) (1 - x) - (1 - x) = 0

x^{4} = \frac{1}{2}

30 x^{2} - 27 x - 3 x^{3} = 0

What are the solutions to the equation z^3=2^{5/2}(1+i) ?
The solutions to the equation z^3=2^{5/2}(1+i) are
Find the zeros of z^3=2^{5/2}(1+i)
The zeros of z^3=2^{5/2}(1+i) are