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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

z^2(1+i)=2i

z^{2} (1 + i) = 2 i

z = \frac{- 2 2 - 2 + 2 2 + 1}{2} + \frac{2 - 1}{2} i, z = \frac{2 2 - 2 - 2 2 + 1}{2} - \frac{2 - 1}{2} i

Solution steps

z^{2} (1 + i) = 2 i

Substitute

z = x + y i

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

Expand

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

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

Rewrite

2 i

in standard complex form:

0 + 2 i

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

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

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

[x^{2} - y^{2} - 2 x y = 0 x^{2} - y^{2} + 2 x y = 2] : x = \frac{- 2 2 - 2 + 2 2 + 1}{2}, x = \frac{2 2 - 2 - 2 2 + 1}{2}, y = \frac{2 - 1}{2} y = - \frac{2 - 1}{2}

Substitute back

z = x + y i

z = \frac{- 2 2 - 2 + 2 2 + 1}{2} + \frac{2 - 1}{2} i, z = \frac{2 2 - 2 - 2 2 + 1}{2} - \frac{2 - 1}{2} i

z = z + \frac{i}{z}

s i (x + 3)^{2} = (x - 1)^{2} + 18

x, f s i f (x) = x^{2} - 3 x + 2

x^{2} + 2 - i = 0

x + y i - x i + y = 3