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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Substitute

z = x + y i

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

Expand

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

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

Rewrite

0

in standard complex form:

0 + 0 i

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

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

Substitute back

z = x + y i

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

3 x - 2 y - x i + 3 y = i

z^{2} + z = 11 + 3 i

∣ 2 (a - i y) + i ∣ = 4

x^{2} = \frac{- 1 + i 3}{2}

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

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