What is the answer to z^2+4iz+5=0 ?

The answer to z^2+4iz+5=0 is z=-5i,z=i

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z^2+4iz+5=0

z^{2} + 4 i z + 5 = 0

z = - 5 i, z = i

Solution steps

z^{2} + 4 i z + 5 = 0

Substitute

z = x + y i

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

Expand

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

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

Rewrite

0

in standard complex form:

0 + 0 i

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

[x^{2} - y^{2} - 4 y + 5 = 0 4 x + 2 x y = 0] : (x = 0, x = 0, y = - 5 y = 1)

Substitute back

z = x + y i

z = - 5 i, z = i

\frac{2 + 5 i}{1 - i} = x + y i

a, ∣ a + bi ∣ = 1

a, ∣ a + bi ∣ = 3

i (x) = x^{2} + 4 x + 7

z = 2 e^{i (\frac{π}{3})}