Let w, z be complex numbers. Show that |wz|=|w||z|, and using the fact that x=|x|e^{arg(x)i}, show further that arg(wz)=arg(w)+arg(z) where |.| is the absolute value and arg(.) is the angle (in polar coordinates). Hence, find all solutions to x^n=1 .
Put w=a+bi, z=c+di where a, b, c, d are real, and i2=-1. Then |wz|2=|(a+bi)(c+di)|2=|ac-bd+(ad+cb)i|2=(ac)2+(bd)2-2bcd+(ad)2+(cb)2+2abcd=(ac)2+(bd)2+(ad)2+(cb)2=(a2+b2)(c2+d2)=|w|2|z|2. Taking square roots, we get |wz|=|w||z|, noting that these values can never be negative. For the argument, wz=|w|e^{arg(w)i}|z|e^{arg(z)i}=|w||z|e^{(arg(w)+arg(z))i} so that arg(wz)=arg(w)+arg(z).
Now, suppose xn=1. Write x=|x|e^{arg(x)i}, 1=e^{0i}, then we rewrite the equation as (|x|e^{arg(x)i})n=e^{0i}. Since we are multiplying x to itself n times, we use the result from the previous part to get that xn=|x|ne^{n arg(x)i}. By comparing magnitute and arguments, we see that |x|=1 and arg(x)n=0, or equivalently, arg(x)n=2kpi for some natural number k, since e^{xi} is periodic in x with period 2pi. Notice that 0<=arg(x)<2pi, so that 0<=k<=n. So then arg(x)=2pi k/n for k=0, 1, ..., n, which gives us the n+1 different solutions x=e^{2ipik/n} for k=0, 1, ..., n. However, notice that when k=0, and when k=n gives us the same solution, so infact we only get n distinct solutions x=e^{2ipik/n} for k=0, 1, ..., n-1.
