- #1
Tsunoyukami
- 215
- 11
I'm not entirely sure if my reasoning is correct here...
"[G]ive the order of each of the zeroes of the given function.
6. ##Log(1 - z)##, ##|z| < 1##" (Complex Variables, 2nd edition by Stephen D. Fisher; pg. 133)
The complex logarithm function is defined in the following manner:
##log(z) = ln|z| + iarg(z)##
##Log(z) = ln|z| + iArg(z)##, ##Arg(z) \in [0, 2\pi)## (this is known as the principal branch)
As a result of this definition, the principal logarithm can only have a zero when both ##ln|z| = 0## and ##iArg(z) = 0##. This is true if and only if ##|z| = 1## and ##Arg(z) = 0##; that is to say, for the complex number ##z = x + iy = 1 + i \cdot 0 = 1##. However, there is an issue with this as the problem explicitly states that ##|z| < 1## and so the function given in the question is never zero - therefore this function has no zeroes. Is this correct?My next question is suppose I had a similar function..say ##g(z) = log(z-1)##, ##|z| < 2## - this function would have an infinite number of zeroes (because of the way "little log" is defined) of order one corresponding to each ##z = 1 + 2 \pi in## where n is an integer. Is that correct?
"[G]ive the order of each of the zeroes of the given function.
6. ##Log(1 - z)##, ##|z| < 1##" (Complex Variables, 2nd edition by Stephen D. Fisher; pg. 133)
The complex logarithm function is defined in the following manner:
##log(z) = ln|z| + iarg(z)##
##Log(z) = ln|z| + iArg(z)##, ##Arg(z) \in [0, 2\pi)## (this is known as the principal branch)
As a result of this definition, the principal logarithm can only have a zero when both ##ln|z| = 0## and ##iArg(z) = 0##. This is true if and only if ##|z| = 1## and ##Arg(z) = 0##; that is to say, for the complex number ##z = x + iy = 1 + i \cdot 0 = 1##. However, there is an issue with this as the problem explicitly states that ##|z| < 1## and so the function given in the question is never zero - therefore this function has no zeroes. Is this correct?My next question is suppose I had a similar function..say ##g(z) = log(z-1)##, ##|z| < 2## - this function would have an infinite number of zeroes (because of the way "little log" is defined) of order one corresponding to each ##z = 1 + 2 \pi in## where n is an integer. Is that correct?