# Thread: finding a complex numbers

1. ## finding a complex numbers

Hi folks,

I am asked to use the definition $\cos z = \dfrac{e^{iz} + e^{-iz}}{2}$ to find 2 imaginary numbers having a cosine of 4.

so, $\cos z = 4$ therefore

$4 = \dfrac{e^{iz} + e^{-iz}}{2}$

$8e^{iz} = e^{i2z} + 1$

let $w = e^{iz}$

$w^2 - 8w + 1 = 0$

$w = 4 \pm \sqrt{15}$

$e^{iz} = 4 \pm \sqrt{15}$

$iz \ln e = \ln (4 \pm \sqrt{15})$

multiply both sides by i

$i^2 z = i \ln(4 \pm \sqrt 15)$

$-z = i \ln(4 \pm \sqrt 15)$

$z = i \ln \dfrac{1}{(4 \pm \sqrt 15)}$

$z = i \ln (4 \pm \sqrt{15})$

can anyone see what I am doing wrong?

3. ## Re: finding a complex numbers

Well, that's a first! Me being right and the book wrong! I think I will start a small celebration. Where's that bottle?

4. ## Re: finding a complex numbers

the book is also right

5. ## Re: finding a complex numbers

I did consider this possibility. I thought that it may be due to the plus or minus sign. It could absorb the -z, but I wasn't convinced and sought the wisdom of the experts. I must admit Idea, that I like your minimalist style, but this time, it is too minimal. Could you kindly explain?

6. ## Re: finding a complex numbers

$\displaystyle \left(4+\sqrt{15}\right)\left(4-\sqrt{15}\right)=1$

therefore

$\displaystyle \ln \left(4+\sqrt{15}\right)=\ln \left(\frac{1}{4-\sqrt{15}}\right)$

7. ## Re: finding a complex numbers

Yes. I was wrong about the book being wrong.

8. ## Re: finding a complex numbers

Very neat! Thanks Idea.

9. ## Re: finding a complex numbers

Another way to see this is that $\cos{(x)} = \cos{(-z)}$. This holds in the complex domain too because the MacLaurin series contains only even powers of the variable.

Thus, if $z=z_0$ is an answer, $z=-z_0$ must also be an answer.