Sinhz = 0 iff z=n(pi)i (n=0, +/- 1, +/- 2 ) ~ a question about this?

  • Thread starter laura_a
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Homework Statement


I have this weird quesiton in my text book I'm not even sure which part is the question and what i need to do...

It says

"Give details showing that the zeros of sinhz and coshz are as in statements (14) and (15) "


Homework Equations



(14) sinhz = 0 iff z=n(pi)i (n=0, +/- 1, +/- 2...)

(15) coshz = 0 iff z=(pi/2 + n(pi)) (n=0, +/- 1, +/- 2...)

-sin(iz) = sinh(z)
cos(iz) = cosh(z)
sinh(z) = [e^z - e^(-z) ] /2
cosh(z) = [e^z + e^(-z)]/2

The Attempt at a Solution



I haven't made a start on the quesiton because I don't understand the words and what it wants me to calc? :grumpy:
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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Substitute the given values into your formulas for sinh and cosh and show they are zero. The words look pretty clear to me. I think you've forgotten an i in your cosh zeros.
 

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