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Old 2016-12-06, 10:21   #3
Nick's Avatar
Dec 2012
The Netherlands

25·32·5 Posts

Yes, it's an alarming story!

Bringing a square inside the bracket is a good idea and then, as you say, you get the norm N(1+i) instead of the absolute value. And N(1+i)=(1+i)(1-i) so you can cancel 1+i, getting \(\left(\frac{1+i}{1-i}\right)^{150}\). And we know that 1-i=-i(1+i)...

Another way of approaching this is geometrically.
Let \(z=\frac{1}{\sqrt{2}}(1+i)\). Then (associating each complex number x+yi with the point (x,y) in the plane), z lies on the circle with centre (0,0) and radius 1:
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Using proposition 47, we see that \(z^2\) also lies on that circle but with angle 2t from the x-axis instead of t. Similarly, \(z^3\) lies on the circle with angle 3t to the x-axis, and so on. In particular, \(z^8=1\).
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