What is the answer to 2(a+ib)^2+2(a+ib)+13=0 ?

The answer to 2(a+ib)^2+2(a+ib)+13=0 is

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2(a+ib)^2+2(a+ib)+13=0

2 (a + ib)^{2} + 2 (a + ib) + 13 = 0

(a = - \frac{1}{2}, a = - \frac{1}{2}, b = \frac{5}{2} b = - \frac{5}{2})

Solution steps

2 (a + ib)^{2} + 2 (a + ib) + 13 = 0

Expand

2 (a + ib)^{2} + 2 (a + ib) + 13 : (2 a^{2} - 2 b^{2} + 2 a + 13) + i (2 b + 4 ab)

(2 a^{2} - 2 b^{2} + 2 a + 13) + i (2 b + 4 ab) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(2 a^{2} - 2 b^{2} + 2 a + 13) + i (2 b + 4 ab) = 0 + 0 i

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

[2 a^{2} - 2 b^{2} + 2 a + 13 = 0 2 b + 4 ab = 0]

[2 a^{2} - 2 b^{2} + 2 a + 13 = 0 2 b + 4 ab = 0] : (a = - \frac{1}{2}, a = - \frac{1}{2}, b = \frac{5}{2} b = - \frac{5}{2})

(a = - \frac{1}{2}, a = - \frac{1}{2}, b = \frac{5}{2} b = - \frac{5}{2})

R, ii R + i q = e

∣ z - (3 + 2 i) ∣ = 2

x^{2} - 3 x + i x + 4 = 3 i

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

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

What is the answer to 2(a+ib)^2+2(a+ib)+13=0 ?
The answer to 2(a+ib)^2+2(a+ib)+13=0 is