What is the answer to i(x)=e(2x^2-1) ?

The answer to i(x)=e(2x^2-1) is x=(sqrt(8e^2-1))/(4e)+1/(4e)i,x=-(sqrt(8e^2-1))/(4e)+1/(4e)i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

i(x)=e(2x^2-1)

i (x) = e (2 x^{2} - 1)

x = \frac{8 e ^{2} - 1}{4 e} + \frac{1}{4 e} i, x = - \frac{8 e ^{2} - 1}{4 e} + \frac{1}{4 e} i

Solution steps

i (x) = e (2 x^{2} - 1)

Substitute

x = a + bi

i (a + bi) = e (2 (a + bi)^{2} - 1)

Expand

- b + ia = (2 e a^{2} - 2 e b^{2} - e) + 4 e iab

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

[- b = 2 e a^{2} - 2 e b^{2} - e a = 4 e ab]

[- b = 2 e a^{2} - 2 e b^{2} - e a = 4 e ab] : (a = \frac{8 e ^{2} - 1}{4 e}, a = - \frac{8 e ^{2} - 1}{4 e}, b = \frac{1}{4 e} b = \frac{1}{4 e})

Substitute back

x = a + bi

x = \frac{8 e ^{2} - 1}{4 e} + \frac{1}{4 e} i, x = - \frac{8 e ^{2} - 1}{4 e} + \frac{1}{4 e} i

z \cdot z + z = 3 - i

a + bi = - i

x + 2 i = 5 + y i

x^{2} + i x 3 + i = 0

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

What is the answer to i(x)=e(2x^2-1) ?
The answer to i(x)=e(2x^2-1) is x=(sqrt(8e^2-1))/(4e)+1/(4e)i,x=-(sqrt(8e^2-1))/(4e)+1/(4e)i