What is the answer to z^2-(5+4i)z+20i=0 ?

The answer to z^2-(5+4i)z+20i=0 is z=4i,z=5

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z^2-(5+4i)z+20i=0

z^{2} - (5 + 4 i) z + 20 i = 0

z = 4 i, z = 5

Solution steps

z^{2} - (5 + 4 i) z + 20 i = 0

Substitute

z = x + y i

(x + y i)^{2} - (5 + 4 i) (x + y i) + 20 i = 0

Expand

(x + y i)^{2} - (5 + 4 i) (x + y i) + 20 i : (x^{2} - y^{2} - 5 x + 4 y) + i (20 - 4 x - 5 y + 2 x y)

(x^{2} - y^{2} - 5 x + 4 y) + i (20 - 4 x - 5 y + 2 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(x^{2} - y^{2} - 5 x + 4 y) + i (20 - 4 x - 5 y + 2 x y) = 0 + 0 i

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

[x^{2} - y^{2} - 5 x + 4 y = 0 20 - 4 x - 5 y + 2 x y = 0]

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

Substitute back

z = x + y i

z = 4 i, z = 5

4 (x - y i) = 8 + 2 y i + 10 i

(1 + 2 i) x + (3 - 5 i) y = 1 - 3 i

r (t) = t^{3} i + t^{2} j

z^{2} = - 2 + 23 i

x^{2} = 1 + i

What is the answer to z^2-(5+4i)z+20i=0 ?
The answer to z^2-(5+4i)z+20i=0 is z=4i,z=5