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I got a question I am unable to work out. Since it's quadratic with complex coefficients I believe we should use the quadratic equation but it doesn't seem to work out.

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

Below is what I did:

a = 1
b = -(1+i)
c = 2+2i

After plugging in the quadratic equation it is:

$-(-(1+i))\pm\sqrt{({-(1+i))}^2-4(1)(2+2i)}\over2(1)$

$(-1-i)^2 = 1^2+2i+i^2 = 1 + 2i -1 = 2i$

$(1+i)\pm\sqrt{2i-8-8i)}\over2$

$(1+i)\pm\sqrt{-8-6i)}\over2$

I don't think i can simplify it further after that but the answer should be $2i$ and $1-i$.
Did I make a mistake somewhere?

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    try completing the square – 2017-01-02
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    also note that 2i + (1-i) = 1 + i; 2i * (1-i) = 2+2i. these are vieta's formulas – 2017-01-02
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    Hint: $$(1-3i)^2=-8 - 6i$$ – 2017-01-02
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    ah so that factoring is the key. Thanks a lot leon. – 2017-01-02

2 Answers 2

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Let's continue from where you stopped. Consider

$$-8-6i=1-9-6i=1^2+(3i)^2-2\times 3i=(3i-1)^2$$

And so the roots are

$$\begin{align} x_1=&{1+i+3i-1\over 2}=2i\\x_2=&{1+i-3i+1\over 2}=1-i\end{align}$$

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    I was going to give an elaborate explanation using the arithmetic of the Gaussian numbers, but I like your method much better. +1. – 2017-01-02
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You can't solve $-(1+i)^2 $ in this type.

$(-1-i)^2 = 1^2+2i+i^2 = 1 + 2i -1 = 2i$

Because on solving,

$-(1+i)^2 $ we get $-2i$

$-(1+i)^2 = -(1^2+2i+i^2) = -(1+2i-1) = -2i$

Then,

$(1+i)\pm\sqrt{-(-2i)-8-8i}\over2$

$(1+i)\pm\sqrt{-8-6i}\over2$

$(1+i)\pm \sqrt{9i^2-6i+1}\over2$

$(1+i)\pm \sqrt{(3i-1)^2}\over2$

$(1+i)\pm (3i-1)\over2$

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

On solving you get 2i, 1-2i.

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    why did you put the negative sign outside the square? b includes the negative sign – 2017-01-02
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    I only want to tell you that you can't insert negative sign like that. That's not proper style. – 2017-01-02
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    B=-(1+i) When you insert it into the quadratic equation b^2 becomes (-(1+i))^2. I just simplified it to (-1-i)^2 – 2017-01-02
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    I also don't know where you get the extra - in front of the b^2 – 2017-01-02