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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

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