Graph for the imaginary part of complex logarithmic exponential function

When writing the following in Mathlab:

t = 0:pi/100:pi/2;
r = 0.5:0.1:1;
x = r' * cos(t);
y = r' * sin(t);
z = x + i*y;
w = log(z.^12);
surfc(x, y, imag(w))

We get a graph that shows three "leaps". How can I find out what is the value of t at each "leap"?

[John D'Errico]

That is funny. I just tried your code. No "leaps" that I saw. Are you SURE there are leaps in that graph? This is not even a leap year. Test your code.

[Zed]

The graph will show a quarter of a circle ring with three leaps, one that looks to be at pi/4. Try again.

Answer 1

[Roger Stafford]

For any combination of elements of t and r your w can be expressed as

 w(r,t) = r^12*exp(12*t*i)

so its logarithm is

 log(w(r,t)) = 12*log(r) + 12*t*i

However matlab's 'log' function constrains its imaginary part to the interval from -pi to +pi. Therefore you can expect discontinuities to occur at t = 1/12*pi, 3/12*pi, and 5/12*pi.

[Zed]

Thanks! I understand the first discontinuity is at t=1/12*pi, but why isn't the second at t=2/12*pi and the third at t=3/12*pi? I thought the discontinuity would be at every 1/12 of pi.

[Roger Stafford]

When t increases to 1/12*pi, then 12*t reaches pi. A bit beyond that point it has 2*pi subtracted from it, dropping down to -pi, so then its formula becomes 12*t-2*pi. Next t has to get up to 3/12*pi before this reaches pi again: 12*t-2*pi = 12*(3/12*pi)-2*pi = pi, so then it has 2*pi subtracted again with the formula becoming 12*t-4*pi. That remains valid until t reaches 5/12*pi whereupon we get 12*(5/12*pi)-4*pi = pi once again, so beyond that another 2*pi is subtracted giving 12*t-6*pi. That last is still valid when t reaches 1/2*pi, so that gives you four distinct pieces as t ranges from 0 to 1/2*pi separated by three discontinuities at t = 1/12*pi, 3/12*pi, and 5/12*pi.