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I am stuck on the following problem that says:

How many complex numbers $z=x+iy$ are there such that $x+y=1$ and $e^{i(x^2+y^2)}=1.$ The options are as follows:
$2.$Non-zero but finitely many
$3.$Countably infinite
$4.$Uncountably infinite.

My Attempt: From $e^{i(x^2+y^2)}=1= e^{i(2n \pi)}$ which gives $x^2+y^2=2n \pi$ (where $n \in \mathbb N$) that indicates family of concentric circles with center at the origin. The required solution is the intersection of $x^2+y^2=2n \pi$ and the line $x+y=1$.But now I can not draw the conclusion.Am I going in the right direction? Can someone throw light on it.Thanks in advance for your time.

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That's a great start. You can take 4 out of the list. – 1015 Feb 21 '13 at 17:43
You're doing very well... I think you must have found the intersection of these circles with the line. What is holding you up from finishing? – rschwieb Feb 21 '13 at 17:45
Like @rschwieb says, you're almost done. Substitute $y=1-x$ in the circle equations and solve the resulting quadratic equation in $x$. You'll find a neat parametrization of your solution set. – 1015 Feb 21 '13 at 17:48
This is not about the complex number $z$ at all! It should be "How many points $(x,y) \in \mathbb R^2$ are there..." – TonyK Feb 21 '13 at 18:02
up vote 2 down vote accepted

$x+y=1$ describes a line. Any sufficiently big circle around the origin intersects this lline in two points. So we must have countably many solutions.

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Yes, but now consider how many times do the line and circles intersect. Looks like it would be either option 3 or 4 though one is more likely if you consider the domain of n. How many unique values of n exist?

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I was trying to give a hint rather than finish the answer. – JB King Feb 21 '13 at 17:52
Thanks a lot sir.I have got it. I guess option $3$ will be my friend. – learner Feb 21 '13 at 17:55

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