Factorzing polynomials with complex coefficients

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The discussion revolves around factoring the polynomial z^2 - (2i + 4)z + 8i, with the correct factors being (z - 4)(z + 2i). The original poster struggled with the quadratic formula and completing the square, leading to confusion about the roots. A participant pointed out that the square root of a complex number, specifically √(3 - 4i), is essential for finding the correct solutions. After clarification, the original poster expressed gratitude for the guidance and confirmed their understanding of complex square roots. The conversation highlights the importance of mastering complex number operations in polynomial factorization.
FelixHelix
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Not sure if this is the right place to post (but its related to a complex analysis questions)

I'm doing a past paper for my revision and am stuck at the first hurdle. I simply cannot factor this polynomial in z for the life of me. I've tried completing the square and the usual quadratic formula but do not get the answer as given.

I know what the answers are supposed to be so if anyone could help walk me through how you get there and if there is a technique of generally doing so I'd be most grateful.

z^2-(2i+4)z + 8i = (z-4)(z+2i)
 
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Why don't you show us how you tried doing the quadratic formula? Because it should work
 
I can:

z1,z2= \frac{(2i+4) \pm \sqrt{(2i+4)^2-(4)(1)(8i)}}{(2)(1)}
this simplifies to:
z1,z2 = (i+2) \pm \sqrt{3-4i}

Which isn't what I need...

Do you get the solutions z = 4 and z = 2i?
 
Last edited:
You did get those numbers. Do you know what the square root of 3-4i is?
 
Hi Shredder - No I didn't know how to take the sqrt of a complex number... but I do now. Thanks for pointing this out - I looked it up and am happy with dealing with these now. Your help is much appreciated!

Felix
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

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