View Single Post
 2016-12-06, 10:21 #3 Nick     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: 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, $$z^8=1$$.