Roots of a polynomial with a complex coefficient.

I'm trying to teach myself complex analysis (I didn't have it in undergrad, and am doing my master's in France, where they all did have it in undergrad) with a book online, and one of the first exercises is causing me problems, as it isn't covered in the chapter. It's pretty trivial, so I probably should have learned it in some class a long time ago (maybe even in high school??), but I've never actually had to solve an equation like this. Here's the problem:

Find all solutions of $z^2 + 2z + (1-i) = 0$.

The answers given at the end of the book are in exponential form. I don't know many methods solving in this form, as I just learned a simple one from youtube.

Again, I'm sure it's terribly trivial, but if someone could show me how to solve this, that'd be great.

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Excuse the question: where did you study (mathematics, I presume) that you didn't have at least one introductory course in complex analysis in undergraduate level? –  DonAntonio Jul 11 '12 at 14:38
Adding on Anon's comment: remember the quadratic formula applies in every field with characteristic $\,\neq 2\,$ –  DonAntonio Jul 11 '12 at 14:39
Wikipedia: The formula and its derivation remain correct if the coefficients $a$, $b$ and $c$ are complex numbers, or more generally members of any field whose characteristic is not 2. (In a field of characteristic 2, the element $2a$ is zero and it is impossible to divide by it.) The symbol $\pm \sqrt {b^2-4ac}$ in the formula should be understood as "either of the two elements whose square is $b^2 - 4ac$, if such elements exist". –  Martin Sleziak Jul 11 '12 at 15:54
I studied at a well-ranked public university in the midwest. They had complex analysis, but for a bachelor's degree it was a choice, and I didn't take it. I also never had topology (except a tiny bit included in Real Analysis) and taught it to myself in January before my first semester of grad school. In retrospect, I should have taken both, but I didn't realize it would be so important. –  JKH Jul 11 '12 at 19:26
If you found any answer below helpful, mark it as accept. –  Host-website-on-iPage Jun 24 '13 at 7:57

You can complete the square, and in case you knew what $e^z$ behaves like, here's a solution:

$$z^2+2z+(1-i)=z^2+2z+1-i=(z+1)^2-i=0$$ so that $$(z+1)^2=i=e^{\frac{i\pi}2}$$ hence $$z+1=e^{\frac{i\pi}4}\,\,\vee\,\, e^{\frac{5i\pi}4}$$ $$z=1+e^{\frac{i\pi}4}\,\,\vee\,\,1+e^{\frac{5i\pi}4}$$

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As an addendum, it's worth noting that this is the same as $z=1\pm e^{\frac{i\pi}{4}}$. –  Cameron Buie Jul 11 '12 at 14:52
Thank you, I had gotten z= -1 +/- (i)^1/2 from the quadratic formula, but wasn't sure how to put it into exponential form. (I see now). –  JKH Jul 11 '12 at 19:29

Hint $\$ By the quadratic formula $\rm\: z = -1 \pm \sqrt{\it i\,}.\:$ By my Simple Denesting Rule, up to sign,

$$\sqrt{\it i\,}\, =\, \frac{{\it i} - 1}{\sqrt{-2}}\, =\, \alpha\,(1 + {\it i}),\ \ \ \alpha :=\frac{\sqrt{2}}2 =\, 0.7071\ldots$$

Therefore $\rm\: z\, =\, -1\pm\sqrt{{\it i}\,}\,=-1\pm \alpha\,(1 + {\it i})\, =\, -1\pm \alpha\, \pm\, \alpha\,{\it i}$

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That's what I got with the quadratic formula, but I didn't know how to put it into exponential form. Now I see from the previous answer. –  JKH Jul 11 '12 at 19:30
@JKH There's no need to use any high-powered transcendental methods such as exponentials. As I show above one can compute $\,\sqrt{\it i\,}\,$ purely algebraically, viz. $\rm\:\sqrt{{\it i}\,} = (1+{\it i})/\sqrt{2}\ \$ –  Bill Dubuque Jul 11 '12 at 19:36
The answer in the back of the book is in exponential form, i.e. e^(iπ/4), which is why I wanted it in that form. I didn't realize my answer was correct because I didn't know how to put it in the form the answer was given in. –  JKH Jul 11 '12 at 19:47
@JKH Ah, I missed that. In any case, now you know how to do it a couple ways. –  Bill Dubuque Jul 11 '12 at 20:07