Factoring polynomial with complex coefficients

Given the equation $z^2+4iz-13=0$, solve for $z$ without the quadratic formula.

In real numbers set, when I find this kind of equations I usually complete the perfect square trinomial.In this case:

$(z^2+4iz-4)-13+4=0$

$(z+2i)^2-9=0$

I chosen $-4$ because the number whose double is $4i$, is $2i$. And the square of $2i$ is $-4$.

$z+2i= \pm \sqrt{9}$

$z=3-2i \vee z=-3-2i$

Is this correct?Thanks

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Yes. This is correct, and this method demonstrates that you can factor any quadratic polynomial in $z$ by completing the square (the closed form of which is called the quadratic equation). We are using the fact that $AB = 0 \implies A = 0$ or $B = 0$. Such rings where this fact is true are called Integral domains. The set of real numbers and the set of complex numbers are two examples of such integral domains. – JavaMan Feb 9 '12 at 22:26
For fun, let's do it in silly high school style. We want to find two numbers whose product is $13$ and whose sum is $-4i$, It is clear that $3-2i$ and $-3-2i$ work. – André Nicolas Feb 9 '12 at 23:13

Yes it is correct.

You could also have 'simplified' it by setting $w = iz$, and getting a quadratic in $w$ with only real coefficients.

On a side note: Questions which have ill defined things like 'without using quadratic formula', 'without using secant' etc are ridiculous and ought to be banned from classrooms.

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I gave a +1 for the last sentiment, which I strongly agree with... – J. M. Feb 9 '12 at 23:03
@Aryabhata : One should not use the quadratic formula if the quadratic formula is what one is trying to prove. – Michael Hardy Feb 10 '12 at 1:11

What you did is correct.

You can check by substitution: If $z=3-2i$ then

$$(z+2i)^2-9= ((3-2i)+2i)^2-9 = 3^2-9=0.$$

If $z=-3-2i$ then $$(z+2i)^2 - 9 = ((-3-2i) + 2i)^2 - 9 = (-3)^2 - 9 = 0.$$

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