What is the answer to (z-i)^2=3+4i ?

The answer to (z-i)^2=3+4i is z=2+2i,z=-2

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(z-i)^2=3+4i

(z - i)^{2} = 3 + 4 i

z = 2 + 2 i, z = - 2

Solution steps

(z - i)^{2} = 3 + 4 i

Substitute

z = x + y i

(x + y i - i)^{2} = 3 + 4 i

Expand

(x + y i - i)^{2} : (x^{2} - 1 - y^{2} + 2 y) + i (- 2 x + 2 x y)

(x^{2} - 1 - y^{2} + 2 y) + i (- 2 x + 2 x y) = 3 + 4 i

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

[x^{2} - 1 - y^{2} + 2 y = 3 - 2 x + 2 x y = 4]

[x^{2} - 1 - y^{2} + 2 y = 3 - 2 x + 2 x y = 4] : (x = 2, x = - 2, y = 2 y = 0)

Substitute back

z = x + y i

z = 2 + 2 i, z = - 2

x^{2} - 12 i x - 1 = 0

a - bi (a + bi + 1) = 11 + 3 i

\frac{360}{i r} - 1500 = 0

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

1 0^{3} = \frac{i}{i o}