What is the answer to x^2=16i ?

The answer to x^2=16i is x=2sqrt(2)+2sqrt(2)i,x=-2sqrt(2)-2sqrt(2)i

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x^2=16i

x^{2} = 16 i

x = 22 + 22 i, x = - 22 - 22 i

Solution steps

x^{2} = 16 i

Substitute

x = a + bi

(a + bi)^{2} = 16 i

Expand

(a + bi)^{2} : (a^{2} - b^{2}) + 2 iab

(a^{2} - b^{2}) + 2 iab = 16 i

Rewrite

16 i

in standard complex form:

0 + 16 i

(a^{2} - b^{2}) + 2 iab = 0 + 16 i

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[a^{2} - b^{2} = 0 2 ab = 16]

[a^{2} - b^{2} = 0 2 ab = 16] : (a = 22, a = - 22, b = 22 b = - 22)

Substitute back

x = a + bi

x = 22 + 22 i, x = - 22 - 22 i

x^{2} - x - i = 0

∣ z ∣ = 2 - 3 i + z

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

(2 + i) x = y

3 x - 4 y + (6 x + 2 y) i = 5 i

What is the answer to x^2=16i ?
The answer to x^2=16i is x=2sqrt(2)+2sqrt(2)i,x=-2sqrt(2)-2sqrt(2)i