# Thread: Use cos(Z) = (e^[iz] + e^[-iz]) / 2 to solve cos(Z) = 4

1. ## Use cos(Z) = (e^[iz] + e^[-iz]) / 2 to solve cos(Z) = 4

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!

2. ## Re: Complex Numbers - Help

Problem needing assistance: complex numbers

Use definition cos Z =(e[iz] + e[-iz])/2 to find 2 imaginary numbers having a cosine of 4.

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.

3. ## Re: Complex Numbers - Help

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.

4. ## Re: Complex Numbers - Help

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.
$\frac{e^{iz} - e^{-iz}}{2} \, = \, 4$

$e^{iz} \, - \, e^{-iz} \, = \, 8$

$e^{iz} \, - \, \frac{1}{e^{iz}} \, = \, 8$

This reduces to quadratic equation - solve....

5. ## Re: Complex Numbers - Help

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

6. ## Re: Complex Numbers - Help

I don’t know what your course level. But I will answer your question in a basic way.
The complex function $\cos (z) = \cos (x)\cosh (y) - i\sin (x)\sinh (y)$, where $z=x+yi$.
So if $\cos(z)=4$ we must have $\cos (x)\cosh (y)=4$ and $\sin (x)\sinh (y)=0$.
That happens if $y=0$ but that means $\cos (x)=4$ which is impossible.
Thus we must have $x=0$ or a even multiple of $\pi$ which gives $\cosh(y) = 4\,\& \,y = \frac{{8 \pm \sqrt {68} }}{2}$.
$z= i \frac{{8 \pm \sqrt {68} }}{2}$.

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