What is the answer to z^2+(3-i)z+2-6i=0 ?

The answer to z^2+(3-i)z+2-6i=0 is z=2i,z=-3-i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

z^2+(3-i)z+2-6i=0

z^{2} + (3 - i) z + 2 - 6 i = 0

z = 2 i, z = - 3 - i

Solution steps

z^{2} + (3 - i) z + 2 - 6 i = 0

Substitute

z = x + y i

(x + y i)^{2} + (3 - i) (x + y i) + 2 - 6 i = 0

Expand

(x + y i)^{2} + (3 - i) (x + y i) + 2 - 6 i : (x^{2} - y^{2} + 3 x + y + 2) + i (- 6 - x + 3 y + 2 x y)

(x^{2} - y^{2} + 3 x + y + 2) + i (- 6 - x + 3 y + 2 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(x^{2} - y^{2} + 3 x + y + 2) + i (- 6 - 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 + y + 2 = 0 - 6 - x + 3 y + 2 x y = 0]

[x^{2} - y^{2} + 3 x + y + 2 = 0 - 6 - x + 3 y + 2 x y = 0] : (x = 0, x = - 3, y = 2 y = - 1)

Substitute back

z = x + y i

z = 2 i, z = - 3 - i

7 + 3 x - 3 i = 8 x - 6 y i - 8 + 9 i

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

\frac{x + i y}{x - i y} = - i

(a + i) (b - 3 i) = 7 - 11 i

2 (x + y i) = i (3 - 4 i)

What is the answer to z^2+(3-i)z+2-6i=0 ?
The answer to z^2+(3-i)z+2-6i=0 is z=2i,z=-3-i