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Below is a complex quadratic equation I think I have solved.

$z^2-2z+i=0$ gives $z=\frac{2\pm\sqrt{4-4i}}{2}$. This becomes $z=1\pm\sqrt{1-i}$ with some basic algebra.

My question is: am I done here? In prior problems I had to evaluate expressions such as $\sqrt{i-1}$ to find it equivalent to $\sqrt[4]{2}*\operatorname{cis}(3\pi/8)$ and $\sqrt[4]{2}*\operatorname{cis}(11\pi/8)$. Hopefully this question is not as rudimentary as I suspect it to be!

Charles
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    Your expansion (or is it a simplification?) of $\sqrt{1-i,}$ is correct. I for one would prefer to see that, over the unexpanded form, if I were reading a homework, for instance. – Lubin Aug 27 '16 at 01:45
  • Whether you are done or not depends on the intention of the exercise writer or what you need the answer for. As Lubin said above, generally speaking when answers are complex numbers we prefer to see them in the standard format $a+bi$. – anon Aug 27 '16 at 02:40
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    The notation $\sqrt{~}$ should be used only for square roots of non-negative real numbers, as it denotes the non-negative square root of such a number. For the square roots of complex numbers you have no criterion to make a difference between them. – Bernard Aug 27 '16 at 08:45

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Using the quadratic formula is really not the way to go here. You are ending up with a complex number inside a square root, which is problematic. Here is my take on it for the set up: Completing the square is much better: $(z-1)^2=1-i$. If you momentarily consider $z-1=w$, then we have $w^2=1-i$. It seems you are familiar with $cis$, so let's go down that route. $|w^2|=\sqrt{2}$ and argument is $-\pi/4$. So for $w$ we have $\sqrt[4]{2}$ and two new angles: $-\pi/8$ and $7\pi/8$. Use these values for $cis$. Technically you solved it for $w$ and thus for $z-1$. Adding $1$ solves it for $z$. Now it becomes a trig issue to figure out what things like $cis(-\pi/8)$ would be, but that can be done with some well known half angle formula trig identities. It involves a square root within a square root, but maybe you can try to find that out. Hope this helps...

Matt Samuel
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imranfat
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  • I don't see how complex numbers underneath radical signs are problematic: this answer is precisely an explanation of how to evaluate them. – anon Aug 27 '16 at 02:39
  • To the comments above. It's a difference of opinion regarding notation, and that's fine. In my university we were strictly told that $i$ is the imaginary unit for which $i^2=-1$ and if we dared to use the definition $i=\sqrt{-1}$, we were marked wrong. But lots of others don't take notational issues with complex numbers inside radicals. However I humbly remain of the opinion that completing the square is a better way for complex quadratic equations than the quadratic formula. – imranfat Aug 27 '16 at 03:01
  • Aww, come on! Unlike any odd prime order roots, you can take the square root of any complex number with algebra alone. The square root of $a+bi$ is $x+yi$, where $x^2=(r+a)/2, y^2=(r−a)/2, r=\sqrt{a^2+b^2}$. For positive $b$ take the same signs for $x$ and $y$, for negative $b$ take opposite signs, and for zero $b$ there no additional sign specifications are needed. – Oscar Lanzi Aug 27 '16 at 03:06
  • Isn't a notation like $\sqrt{complex}$ becoming an issue in further complex analysis, where it is important for complex numbers to be a field? That's not OT for the post here, but Cameron has elaborated quite extensively here:http://math.stackexchange.com/questions/664962/what-is-the-square-root-of-complex-number-i . For the purpose of solving an equation like the OP's, it doesn't matter and my argument becomes perhaps a moot point... – imranfat Aug 27 '16 at 03:18