Let F(z)=z−iz+i for all complex numbers z=i, and let zn=F(zn−1) for all positive integers n. Given that z0=1371+i, find z2002.
Solution — click to reveal
Iterating F a few times, we get F(F(z))F(F(F(z)))=z−iz+i−iz−iz+i+i=(z+i)−i(z−i)(z+i)+i(z−i)=z+i−zi−1z+i+zi+1=(z−1)(1−i)(z+1)(i+1)=(z−1)⋅2(z+1)(i+1)2=(z−1)⋅2(z+1)(2i)=z−1z+1i,=z−1z+1i−iz−1z+1i+i=z−1z+1−1z−1z+1+1=(z+1)−(z−1)(z+1)+(z−1)=z.Thus, zk+3=zk for all k. Since 2002≡1(mod3), we then have z2002=z1=z0−iz0+i=1/1371/137+2i=1+274i.