- #1
Bashyboy
- 1,421
- 5
Demonstrate that ##|e^{z^2}| \le e^{|z|^2}##
We have at our disposal the theorem which states ##Re(z) \le |z|##. Here is my work:
##e^{|z|^2} \ge e^{(Re(z))^2} \iff## By the theorem stated above.
##e^{|z|^2} \ge e^x##
We note that ##y^2 \ge 0##, and that multiplying by ##-1## will give us ##- y^2 \le 0##; adding ##x^2## to both sides gives us ##x^2 - y^2 \le x^2##. Substituting this in gives us
##e^{|z|^2} \ge e^{x^2 - y^2}##. I calculated ##|e^{z^2}|## and found that it was ##e^{x^2-y^2}##. Therefore,
##e^{|z|^2} \ge |e^{z^2}|##
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Here is the one issue I see with the proof, but I may have resolved this issue: was the first step justly done? I believe so, and here is why:
##e^{f(z)} \ge e^{g(z)} \iff##
##\ln e^{f(z)} \ge \ln e^{g(z)} \iff##
##f(z) \ge g(z)##.
So, one exponential function is greater than the other when its argument function is greater than the other for all ##z##.
Does this seem correct?
We have at our disposal the theorem which states ##Re(z) \le |z|##. Here is my work:
##e^{|z|^2} \ge e^{(Re(z))^2} \iff## By the theorem stated above.
##e^{|z|^2} \ge e^x##
We note that ##y^2 \ge 0##, and that multiplying by ##-1## will give us ##- y^2 \le 0##; adding ##x^2## to both sides gives us ##x^2 - y^2 \le x^2##. Substituting this in gives us
##e^{|z|^2} \ge e^{x^2 - y^2}##. I calculated ##|e^{z^2}|## and found that it was ##e^{x^2-y^2}##. Therefore,
##e^{|z|^2} \ge |e^{z^2}|##
________________________________________________________________________
Here is the one issue I see with the proof, but I may have resolved this issue: was the first step justly done? I believe so, and here is why:
##e^{f(z)} \ge e^{g(z)} \iff##
##\ln e^{f(z)} \ge \ln e^{g(z)} \iff##
##f(z) \ge g(z)##.
So, one exponential function is greater than the other when its argument function is greater than the other for all ##z##.
Does this seem correct?