What is the answer to x^2-(1+2i)x+(2+i)=0 ?

The answer to x^2-(1+2i)x+(2+i)=0 is x= 1/2-(-2+sqrt(11))/2 i,x= 1/2+(2+sqrt(11))/2 i

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x^2-(1+2i)x+(2+i)=0

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

x = \frac{1}{2} - \frac{- 2 + 11}{2} i, x = \frac{1}{2} + \frac{2 + 11}{2} i

Solution steps

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

Substitute

x = a + bi

(a + bi)^{2} - (1 + 2 i) (a + bi) + 2 + i = 0

Expand

(a + bi)^{2} - (1 + 2 i) (a + bi) + 2 + i : (a^{2} - b^{2} - a + 2 b + 2) + i (1 - b - 2 a + 2 ab)

(a^{2} - b^{2} - a + 2 b + 2) + i (1 - b - 2 a + 2 ab) = 0

Rewrite

0

in standard complex form:

0 + 0 i

(a^{2} - b^{2} - a + 2 b + 2) + i (1 - b - 2 a + 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} - a + 2 b + 2 = 0 1 - b - 2 a + 2 ab = 0]

[a^{2} - b^{2} - a + 2 b + 2 = 0 1 - b - 2 a + 2 ab = 0] : (a = \frac{1}{2}, a = \frac{1}{2}, b = - \frac{- 2 + 11}{2} b = \frac{2 + 11}{2})

Substitute back

x = a + bi

x = \frac{1}{2} - \frac{- 2 + 11}{2} i, x = \frac{1}{2} + \frac{2 + 11}{2} i

\frac{5 + 2 i}{z} = 2 - i

\frac{( x + i y )}{( x - i y )} = - i

(x - y) + 5 i = 2 + y i

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

(u - i) (4 + i v) = 11 + 2 i

What is the answer to x^2-(1+2i)x+(2+i)=0 ?
The answer to x^2-(1+2i)x+(2+i)=0 is x= 1/2-(-2+sqrt(11))/2 i,x= 1/2+(2+sqrt(11))/2 i