In my class, my instructor told us that the square root of a complex number is in general not a function because it is multi-valued. For example, e^(ipi/4) could have a square root of e^(ipi/8) or e^(i*9pi/8). He then added that if we shortened the domain of the polar angle to (-pi,pi), the square root then becomes a function. I don't see how this works. The square root has a period of pi, over which it repeats itself. For eq, both e^(-ipi/4) and e^(ipi/4) qualify as square roots of e^(i*pi/2), and they both lie in the shortened domain. I don't see where I am going wrong.
Subscribe to:
Post Comments (Atom)
analysis - Injection, making bijection
I have injection $f \colon A \rightarrow B$ and I want to get bijection. Can I just resting codomain to $f(A)$? I know that every function i...
-
I need to give an explicit bijection between $(0, 1]$ and $[0,1]$ and I'm wondering if my bijection/proof is correct. Using the hint tha...
-
So if I have a matrix and I put it into RREF and keep track of the row operations, I can then write it as a product of elementary matrices. ...
-
Recently I took a test where I was given these two limits to evaluate: $\lim_\limits{h \to 0}\frac{\sin(x+h)-\sin{(x)}}{h}$ and $\lim_\limi...
No comments:
Post a Comment