What is the answer to (a-2i)^2=21-20i ?

The answer to (a-2i)^2=21-20i is a=5,a=-5+4i

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(a-2i)^2=21-20i

(a - 2 i)^{2} = 21 - 20 i

a = 5, a = - 5 + 4 i

Solution steps

(a - 2 i)^{2} = 21 - 20 i

Substitute

a = x + y i

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

Expand

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

(x^{2} - 4 - y^{2} + 4 y) + i (- 4 x + 2 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} - 4 - y^{2} + 4 y = 21 - 4 x + 2 x y = - 20]

[x^{2} - 4 - y^{2} + 4 y = 21 - 4 x + 2 x y = - 20] : (x = 5, x = - 5, y = 0 y = 4)

Substitute back

a = x + y i

a = 5, a = - 5 + 4 i

x, i = k + \frac{b - x}{x}

\frac{3 b - 2 ai}{4 - 3 i} = 1

(2 + i) x^{2} + (3 - i) x + 1 - 3 i = 0

(x - y) + 3 i = 7 + y i

(3 + 3 i) (x + y i) = 27 - 3 i

What is the answer to (a-2i)^2=21-20i ?
The answer to (a-2i)^2=21-20i is a=5,a=-5+4i