What is the answer to z^2=-15+8i ?

The answer to z^2=-15+8i is z=1+4i,z=-1-4i

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z^2=-15+8i

z^{2} = - 15 + 8 i

z = 1 + 4 i, z = - 1 - 4 i

Solution steps

z^{2} = - 15 + 8 i

Substitute

z = x + y i

(x + y i)^{2} = - 15 + 8 i

Expand

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

(x^{2} - y^{2}) + 2 i x y = - 15 + 8 i

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

[x^{2} - y^{2} = - 15 2 x y = 8]

[x^{2} - y^{2} = - 15 2 x y = 8] : (x = 1, x = - 1, y = 4 y = - 4)

Substitute back

z = x + y i

z = 1 + 4 i, z = - 1 - 4 i

\frac{( - 6 i + n - 3 n + 5 )}{4 + n ^{2}} = 0

(a + 2 i) (1 + 2 i) = 4 - bi

z^{2} - \frac{2 i + 1}{2 - i} = 0

∣ x + y i ∣ = y - i x

\frac{( z - 2 i )}{( z + 2 )} = 4 i