What is the answer to i(x)=ex^2+2pi ?

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

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i(x)=ex^2+2pi

i (x) = e x^{2} + 2 π

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

Solution steps

i (x) = e x^{2} + 2 π

Substitute

x = a + bi

i (a + bi) = e (a + bi)^{2} + 2 π

Expand

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

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

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

[- b = e a^{2} - e b^{2} + 2 π a = 2 e ab] : (a = 0, a = 0, b = - \frac{- 1 + 1 + 8 e π}{2 e} b = \frac{8 e π + 1 + 1}{2 e})

Substitute back

x = a + bi

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

(1 - i) x + 3 y = 2 - 1

a, a + bi = a + c i

(2 x + 12 y) + (x + y) i = 6 - 2 i

(a - 2 bi) - (b - ai) = 2 + i

a + bi = i

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