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

The answer to z^2-(3+3i)z+(1+3i)=0 is z=0.86644…+0.31619…i,z=2.13355…+2.68380…i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

z = 0.86644 \dots + 0.31619 \dots i, z = 2.13355 \dots + 2.68380 \dots i

Solution steps

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

Substitute

z = x + y i

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

Expand

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

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

Rewrite

0

in standard complex form:

0 + 0 i

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

[x^{2} - y^{2} - 3 x + 3 y + 1 = 0 3 - 3 x - 3 y + 2 x y = 0] : (x = 0.86644 \dots, x = 2.13355 \dots, y = 0.31619 \dots y = 2.68380 \dots)

Substitute back

z = x + y i

z = 0.86644 \dots + 0.31619 \dots i, z = 2.13355 \dots + 2.68380 \dots i

(2 x - y) + (x + y) i = - 4 - 5 i

2 + x i = y - 7 i

z^{2} = 7 - 24 i

- \frac{1}{z - i} - 2 i - z = 0

x, g [- 6] i f g [x] = 10 x + 9

What is the answer to z^2-(3+3i)z+(1+3i)=0 ?
The answer to z^2-(3+3i)z+(1+3i)=0 is z=0.86644…+0.31619…i,z=2.13355…+2.68380…i