4 ms·
z = x + i*y = e^(a + i*b) = e^a*(e^(i*b)) = e^a(cos(b)+i*sin(b)) Both sin and cos are many to one functions. In the equation above, replacing b with b +
by NAFV_P 12y ago
z = x + i*y = e^(a + i*b) = e^a*(e^(i*b)) = e^a(cos(b)+i*sin(b))
Both sin and cos are many to one functions. In the equation above, replacing
b
with
b + 2*pi*t
where t is any positive or negative integer, would result in the same complex number.
- NaNaN 12y agoWe've used the same number for those different rotations. That does not do anything directly about irrational power p. If p is a rational number represented by m/n, then 360 / (m/n) * m = 360n So there are multiples of 360 that are divisible by m/n, and we can get a unique complex number from z^p. Now p is irrational, what happens? According to what you said, there are so many different graphs that can not be merged into just one.