What is the answer to (2-i)z^2+(8i-6)z+(4-7i)=0 ?

The answer to (2-i)z^2+(8i-6)z+(4-7i)=0 is z=3-2i,z=1

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(2-i)z^2+(8i-6)z+(4-7i)=0

(2 - i) z^{2} + (8 i - 6) z + (4 - 7 i) = 0

z = 3 - 2 i, z = 1

Solution steps

(2 - i) z^{2} + (8 i - 6) z + (4 - 7 i) = 0

Substitute

z = x + y i

(2 - i) (x + y i)^{2} + (8 i - 6) (x + y i) + 4 - 7 i = 0

Expand

(2 - i) (x + y i)^{2} + (8 i - 6) (x + y i) + 4 - 7 i : (2 x^{2} - 2 y^{2} + 2 x y - 6 x - 8 y + 4) + i (- 7 - x^{2} + y^{2} - 6 y + 8 x + 4 x y)

(2 x^{2} - 2 y^{2} + 2 x y - 6 x - 8 y + 4) + i (- 7 - x^{2} + y^{2} - 6 y + 8 x + 4 x y) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(2 x^{2} - 2 y^{2} + 2 x y - 6 x - 8 y + 4) + i (- 7 - x^{2} + y^{2} - 6 y + 8 x + 4 x y) = 0 + 0 i

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

[2 x^{2} - 2 y^{2} + 2 x y - 6 x - 8 y + 4 = 0 - 7 - x^{2} + y^{2} - 6 y + 8 x + 4 x y = 0]

[2 x^{2} - 2 y^{2} + 2 x y - 6 x - 8 y + 4 = 0 - 7 - x^{2} + y^{2} - 6 y + 8 x + 4 x y = 0] : (x = 3, x = 1, y = - 2 y = 0)

Substitute back

z = x + y i

z = 3 - 2 i, z = 1

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

m^{2} = - aji

(a + bi)^{2} - 2 (a - bi) = 3

y = c^{1} A i (x) + c^{2} B i (x)

(x + y) + (x - y) \cdot i = 6 + 9 \cdot i

What is the answer to (2-i)z^2+(8i-6)z+(4-7i)=0 ?
The answer to (2-i)z^2+(8i-6)z+(4-7i)=0 is z=3-2i,z=1