What is the answer to z^2=9i ?

The answer to z^2=9i is z= 3/(sqrt(2))+3/(sqrt(2))i,z=-3/(sqrt(2))-3/(sqrt(2))i

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z^2=9i

z^{2} = 9 i

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

Solution steps

z^{2} = 9 i

Substitute

z = x + y i

(x + y i)^{2} = 9 i

Expand

(x + y i)^{2} : (x^{2} - y^{2}) + 2 i x y

(x^{2} - y^{2}) + 2 i x y = 9 i

Rewrite

9 i

in standard complex form:

0 + 9 i

(x^{2} - y^{2}) + 2 i x y = 0 + 9 i

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

[x^{2} - y^{2} = 0 2 x y = 9]

[x^{2} - y^{2} = 0 2 x y = 9] : (x = \frac{3}{2}, x = - \frac{3}{2}, y = \frac{3}{2} y = - \frac{3}{2})

Substitute back

z = x + y i

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

a + bi = 4

x^{2} = - 5 + 12 i

\frac{( x + y i )}{( x - y i )} = x - y i

\frac{100}{8 + 6 i} x + (1 - i)^{3} y = 7

z^{2} + 16 i = 0

What is the answer to z^2=9i ?
The answer to z^2=9i is z= 3/(sqrt(2))+3/(sqrt(2))i,z=-3/(sqrt(2))-3/(sqrt(2))i