Real-Variable Expressions of cosh(z) & Finding Complex Zeroes

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The discussion focuses on expressing cosh(z) in terms of real variables and identifying its complex zeroes. The user initially derives cosh(z) as cos(x)cos(y) - sin(x)sin(y) but is corrected that this expression does not yield real numbers for all complex z. The correct formulation leads to cosh(z) = sinh(x)cos(y) - i cosh(x)sin(y), which is zero when sinh(x)cos(y) = 0 and cosh(x)sin(y) = 0. The solutions indicate that cosh(z) = 0 at z = (2k+1)iπ/2, where k is an integer, and the user seeks guidance on accurately representing these zeroes on an Argand diagram. The discussion emphasizes the importance of correctly interpreting the complex function and plotting its zeroes.
fredrick08
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Homework Statement


express cosh(z) in terms of functions with real variable only, hence find all complex zeroes and mark on argand diagram.


The Attempt at a Solution


I did cosh(z)=cosh(x+iy) then got down to cosh(z)=cos(x)cos(y)-sin(x)sin(y)... can anyone confirm this please? also i got told, that cosh(z)=0, when z=pi*k*i+pi*i/2...when k is a interger... I have no idea, how my friend got this, of if its even correct, or what i have done is correct either... but i would appreciate any help, especially with drawing the zeroes on the argand diagram, becasue with the answer i have, there is only 2 per 2pi
 
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Well, what you have can't be right because cosh(z) is NOT a real number for all complex z, but cos(x)cos(y)-sin(x)sin(y) is real for all real x and y.

cosh(x+iy)= \frac{e^{x+iy}+ e^{-x-iy}}{2}= \frac{e^xe^{iy}+ e^{-x}e^{-iy}}{2}
= \frac{e^xe^{iy}+ e^xe^{-iy}- e^xe^{-iy}+ e^{-x}e^{-iy}}{2}
= e^x\frac{e^{iy}+ e^{-iy}}{2}- e^{iy}\frac{e^x+ e^{-x}}{2}
= e^x cos(y)- cosh(x)e^{-iy}

Of course, e^{ix}= cos(x)+ i sin(x) so e^{-iy}= cos(y)- i sin(y).

That is, cosh(z)= e^xcos(y)- cosh(x)(cos(y)- i sin(y))= [e^x- cosh(x)]cos(y)- cosh(x)cos(y)]- i cosh(x)sin(y).

since cosh(x)= (e^x+ e^{-x})/2,
e^x- cosh(x)= e^x- (e^x+e^{-x})/2= (e^x- e^{-x})/2= sinh(x)

cosh(z)= sinh(x)cos(y)- i cosh(x)sin(y)

That will be 0 when sinh(x)cos(y)= 0 and cosh(x)sin(y)= 0.
 
ahh i see what i did wrong... i forgot about the ix, and converted the wrong terms... ok.. so buy solving those 2 equations for zero, give x=0 and y=(2k+1)pi/2 and second eq gives, y=k*pi, and x=(2k+1)i*pi/2 where k is a integer. is that right... but how do i put that on an argand diagram?
 
so cosh(z)=0, when z=(2k+1)i*pi/2... coz that would mean cos((2k+1)pi/2)... which give zero for all half pi's? so is my diagram just a point at iy=I*pi.2 and -I*pi/2.. for the principal values? or is there special way of drawing it?
 
please can anyone help?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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