Problem needing assistance: complex numbers
Use definition cos Z =(e[iz] + e[-iz])/2 to find 2 imaginary numbers having a cosine of 4.
Please note that the iz and the -iz are exponents.
This is for the class of IB Higher Level Math 3.
Thanks!
Problem needing assistance: complex numbers
Use definition cos Z =(e[iz] + e[-iz])/2 to find 2 imaginary numbers having a cosine of 4.
Please note that the iz and the -iz are exponents.
This is for the class of IB Higher Level Math 3.
Thanks!
Please share with us your work - indicating exactly where you are stuck - so that we know where to begin to help you.Originally Posted by adele.fielding
“... mathematics is only the art of saying the same thing in different words” - B. Russell
replaced cos z by 4, expanded the exponents for eto iz = cos z + i sin z took us in circles back to cos z = 4.
Not getting anywhere.
[tex]\frac{e^{iz} - e^{-iz}}{2} \, = \, 4[/tex]Originally Posted by adele.fielding
[tex]e^{iz} \, - \, e^{-iz} \, = \, 8[/tex]
[tex]e^{iz} \, - \, \frac{1}{e^{iz}} \, = \, 8[/tex]
This reduces to quadratic equation - solve....
“... mathematics is only the art of saying the same thing in different words” - B. Russell
Thankyou so much, that was all we needed to complete the problem.
Can you also explain (or provide online reference) how it is that the cos z can have a value of 4? In real numbers, cos x is between plus and minus 1. How should we be thinking about cos z in complex plane?
Thanks,
aF
I don’t know what your course level. But I will answer your question in a basic way.
The complex function [tex]\cos (z) = \cos (x)\cosh (y) - i\sin (x)\sinh (y)[/tex], where [tex]z=x+yi[/tex].
So if [tex]\cos(z)=4[/tex] we must have [tex]\cos (x)\cosh (y)=4[/tex] and [tex]\sin (x)\sinh (y)=0[/tex].
That happens if [tex]y=0[/tex] but that means [tex]\cos (x)=4[/tex] which is impossible.
Thus we must have [tex]x=0[/tex] or a even multiple of [tex]\pi[/tex] which gives [tex]\cosh(y) = 4\,\& \,y = \frac{{8 \pm \sqrt {68} }}{2}[/tex].
[tex]z= i \frac{{8 \pm \sqrt {68} }}{2}[/tex].
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