What is the answer to (z^1)^2=(2-5i)^2 ?

The answer to (z^1)^2=(2-5i)^2 is z=2-5i,z=-2+5i

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(z^1)^2=(2-5i)^2

(z^{1})^{2} = (2 - 5 i)^{2}

z = 2 - 5 i, z = - 2 + 5 i

Solution steps

(z^{1})^{2} = (2 - 5 i)^{2}

Substitute

z = x + y i

((x + y i)^{1})^{2} = (2 - 5 i)^{2}

Expand

(x^{2} - y^{2}) + 2 i x y = - 21 - 20 i

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

[x^{2} - y^{2} = - 21 2 x y = - 20]

[x^{2} - y^{2} = - 21 2 x y = - 20] : (x = 2, x = - 2, y = - 5 y = 5)

Substitute back

z = x + y i

z = 2 - 5 i, z = - 2 + 5 i

a + bi = - 9 + 4 i

a + ib = 1 - (a - ib) + 3 i

\frac{z - 1}{1 + z} = i + 1

x + i y = ∣ x - y i ∣

(1 + i) z^{2} = - 1 + 7

What is the answer to (z^1)^2=(2-5i)^2 ?
The answer to (z^1)^2=(2-5i)^2 is z=2-5i,z=-2+5i