What is the answer to iz^2-(2+3i)z+1+5i=0 ?

The answer to iz^2-(2+3i)z+1+5i=0 is z=1+i,z=2-3i

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iz^2-(2+3i)z+1+5i=0

i z^{2} - (2 + 3 i) z + 1 + 5 i = 0

z = 1 + i, z = 2 - 3 i

Solution steps

i z^{2} - (2 + 3 i) z + 1 + 5 i = 0

Substitute

z = x + y i

i (x + y i)^{2} - (2 + 3 i) (x + y i) + 1 + 5 i = 0

Expand

i (x + y i)^{2} - (2 + 3 i) (x + y i) + 1 + 5 i : (- 2 x y - 2 x + 3 y + 1) + i (5 + x^{2} - y^{2} - 2 y - 3 x)

(- 2 x y - 2 x + 3 y + 1) + i (5 + x^{2} - y^{2} - 2 y - 3 x) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(- 2 x y - 2 x + 3 y + 1) + i (5 + x^{2} - y^{2} - 2 y - 3 x) = 0 + 0 i

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

[- 2 x y - 2 x + 3 y + 1 = 0 5 + x^{2} - y^{2} - 2 y - 3 x = 0]

[- 2 x y - 2 x + 3 y + 1 = 0 5 + x^{2} - y^{2} - 2 y - 3 x = 0] : (x = 1, x = 2, y = 1 y = - 3)

Substitute back

z = x + y i

z = 1 + i, z = 2 - 3 i

(x + y i)^{2} = 5 + 12 i

z^{2} (3 - i) + 5 (1 - i) (1 + i) + 2 + 16 i = 0

y, s i x^{2} + y^{2} = z^{2}

\frac{( 2 - 3 i )}{2 x + y i} = 3 + 2 i

(a + ib)^{2} = 16 + 30 i

What is the answer to iz^2-(2+3i)z+1+5i=0 ?
The answer to iz^2-(2+3i)z+1+5i=0 is z=1+i,z=2-3i