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How can you solve this? $z^2+2(1+i)z=2+2(\sqrt{3}-1)i$

I have tried to compare left and right side with real and imaginary part i then get

$ x^2+2x-y^2-2y=2$

$xy+x+y=(\sqrt{3}-1)$

But this equation can not be solved.

What else can i do? Setting in $z=re^{i\theta}$ does not help either

The answer is: $ \sqrt{3}-1$ and $-(1+\sqrt{3})-2i$

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  • $\begingroup$ "this equation can not be solved": I am not so sure. From the second equation you can draw $x$ or $y$ in terms of the other unknown and plug in the first, giving a univarite polynomial equation. $\endgroup$ – Yves Daoust May 19 '16 at 14:32
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    $\begingroup$ Anyway, you are just asked to apply the quadratic roots formula on complex coefficients. $\endgroup$ – Yves Daoust May 19 '16 at 14:33
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What about the high school formula for the roots of a quadratic? We have

$$z^2+\left[2(1+i)\right]z-2\left(1+(\sqrt3-1)\right)i=0\;\implies\;$$

$$\Delta =(2(1+i))^2+4\left[2\left(1+(\sqrt3-1)\right)\right]=4\left(2i+2+2\sqrt3-2\right)=8\left(\sqrt3+i\right)$$

Continue on and observe that above you almost has the square root of $\;i\;$ ...

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By setting $z^2 + 2(1+i)z-2-2(\sqrt{3}-1)i = 0$, you can see this is a second degree polynomial, we can solve this by using the standard formula for the solution of second degree polynomials (which is also valid in $\mathbb{C}$). You will have to find two roots of the discriminant, and this one can indeed do by using polar coordinates(or maybe things will turn out more nicely and you can see the square root inmediately).

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It's a quadratic, use the quadratic formula $$z = \frac{ -2(1+i) \pm \sqrt{(2(1+i))^2+4(2+2(\sqrt{3}-1)i)}}{2}$$ Now you have to simplify.

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solving $$ z^2+2(1+i)z-2-2(\sqrt{3}-1)i=0 $$ we have: $$ z=-1-i\pm\sqrt{(1+i)^2+2+2(\sqrt{3}-1)i}=-1-i\pm\sqrt{2}\sqrt{1+\sqrt{3}i} $$ now: $1+\sqrt{3}i=2e^{i\pi/3}$ and $$ \sqrt{1+\sqrt{3}i}=\sqrt{2}e^{i\pi/6}=\sqrt{2}\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i \right) $$ so: $$ z=-1-i\pm 2\left(\frac{\sqrt{3}}{2}+\frac{1}{2}i \right)=-1-i\pm (\sqrt{3}+i) $$

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