What is the answer to x^2+4-3i=0 ?

The answer to x^2+4-3i=0 is x= 1/(sqrt(2))+3/(sqrt(2))i,x=-1/(sqrt(2))-3/(sqrt(2))i

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x^2+4-3i=0

x^{2} + 4 - 3 i = 0

x = \frac{1}{2} + \frac{3}{2} i, x = - \frac{1}{2} - \frac{3}{2} i

Solution steps

x^{2} + 4 - 3 i = 0

Substitute

x = a + bi

(a + bi)^{2} + 4 - 3 i = 0

Expand

(a + bi)^{2} + 4 - 3 i : (a^{2} - b^{2} + 4) + i (- 3 + 2 ab)

(a^{2} - b^{2} + 4) + i (- 3 + 2 ab) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(a^{2} - b^{2} + 4) + i (- 3 + 2 ab) = 0 + 0 i

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

[a^{2} - b^{2} + 4 = 0 - 3 + 2 ab = 0]

[a^{2} - b^{2} + 4 = 0 - 3 + 2 ab = 0] : (a = \frac{1}{2}, a = - \frac{1}{2}, b = \frac{3}{2} b = - \frac{3}{2})

Substitute back

x = a + bi

x = \frac{1}{2} + \frac{3}{2} i, x = - \frac{1}{2} - \frac{3}{2} i

- 3 ai - (- 1 - i) b = 3 a - 2

i (x) = \frac{1}{x}

x^{2} = - i

z^{2} + 2 z + 1 + i = 0

z^{- 1} = \frac{1}{3 + 2 i}

What is the answer to x^2+4-3i=0 ?
The answer to x^2+4-3i=0 is x= 1/(sqrt(2))+3/(sqrt(2))i,x=-1/(sqrt(2))-3/(sqrt(2))i