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)
real analysis - How to find $lim_{hrightarrow 0}frac{sin(ha)}{h}$
How to find $\lim_{h\rightarrow 0}\frac{\sin(ha)}{h}$ without lhopital rule? I know when I use lhopital I easy get $$ \lim_{h\rightarrow 0}...
-
I'm just learning how to test series for convergence and have encountered this series from the Demidovich's book and I can't rea...
-
Ok, according to some notes I have, the following is true for a random variable $X$ that can only take on positive values, i.e $P(X $\int_0^...
-
Make a bijection that shows $|\mathbb C| = |\mathbb R| $ First I thought of dividing the complex numbers in the real parts and the c...
No comments:
Post a Comment