What is the answer to (x+iy)^2+1=1 ?

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

(x+iy)^2+1=1

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

(x = 0, y = 0)

Solution steps

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

Expand

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

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

Rewrite

1

in standard complex form:

1 + 0 i

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

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

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

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

(x = 0, y = 0)

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

(2 - 2 i) x^{2} - (11 + 9 i) x - 16 + 6 i = 0

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

(1 + i)^{2} - a (1 + i) + b = 0

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