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I am trying to solve the following equation: $z^2 - (2 + 2i)z - 5 -10i = 0$?

My attempts so far has been trying to complete the square of $z^2 - (2 + 2i)z - 5 -10i$ but I have had no real progress with that. How would you approach this kind of problem?

Thank you very much for your help!

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3 Answers 3

up vote 2 down vote accepted

Alternatively, by completing the square: $$\begin{aligned}z^2-(2+2i)z - 5 - 10i &= z^2-2(1+i)z - 5 - 10i\\&= z^2-2(1+i)z +(i+1)^2 -(i+1)^2 - 5 - 10i\\&= \left(z-(1+i)\right)^2 - (i+1)^2 - 5 - 10i\\&= (z-(1+i))^2 -5-12i\end{aligned}$$

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The discriminant of the equation is $(2+2i)^2-4\cdot1\cdot\{-(5+10i)\}=20+8i=4(5+12i)$

Let $\sqrt{5+12i}=a+ib\implies(a+ib)^2=5+12i$

Equating the real & the imaginary parts, $a^2-b^2=5,2ab=12$ so, $(a^2+b^2)^2=(a^2-b^2)^2+(2ab)^2=13^2\implies a^2+b^2=13$

So, $a^2=9\implies a=\pm3,b=\pm2$

So, $\sqrt{5+12i}=\pm(3+2i)$

So, $z=\frac{2+2i\pm2(3+2i)}{2\cdot1}$

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@Lucas, I think the calculation of the square root of the discriminant (as it is of the form $a+ib$ ) is what makes it different from Quadratic Equation with real coefficients. –  lab bhattacharjee Feb 20 '13 at 9:52

Hint: Try using the quadratic formula $$ z_{1,2}=\frac{-b\pm \sqrt{b^2-4c}}{2}=\frac{2+2i\pm \sqrt{(2+2i)^2+4(5+10i)}}{2}=\cdots $$

Edit: The quadratic formula comes from completing a square, here how it goes in this case: $$ z^2-2(1+i)z-5(1+2i)= (z-(1+i))^2 -(1+i)^2-5(1+2i)=0 $$ Moving the last two guys to the right gives that the solutions are $$ z_{1,2}-(1+i)=\pm \sqrt{(1+i)^2+5(1+2i)} $$ which is the same as the equation in the hint.

BTW: I do not do the actual calculations on purpose.

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If the quadratic formula works, you should be able to "complete the square" of $z^2 - (2 + 2i)z - 5 -10i$ as well? Thank you for your hint! –  Lukas Arvidsson Feb 20 '13 at 9:28
    
Of course, the quadratic formula comes from completing a square! –  Lior B-S Feb 20 '13 at 9:30
    
Yes that I what I thought as well @Lior B-S. But I seem to lack a strategy when trying to do complete the square of this equation –  Lukas Arvidsson Feb 20 '13 at 9:34
    
Added it to the answer –  Lior B-S Feb 20 '13 at 9:36
    
Thank you very much! –  Lukas Arvidsson Feb 20 '13 at 9:36

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