Writing cos(π+i) in Cartesian form?

Question

I have a problem writing cos(π+i) in Cartesian form (x + yi). I understand that it is usually written in the form cos(θ) + i sin(θ), but I don't understand how to obtain the sin part (and how to even write this in any other form e.g polar form).
any help would be appreciated!

@Kuba just a math problem I have to solve, "Write the following complex numbers in Cartesian form". — , Oct 09 '17 at 17:59
Use the formulas and explanation here: http://www.milefoot.com/math/complex/functionsofi.htm — Srini, Oct 09 '17 at 18:07

score 3 · Accepted Answer · answered Oct 09 '17 at 18:16

3

$\cos(i+\pi)=-\cos(i)=-\cosh(1)$

answered Oct 09 '17 at 18:16

gammatester

18,827

Thank you, I guess there's no way around hyperbolic functions with this one (although by syllabus does not mention these anywhere). – Thijs H Oct 09 '17 at 18:22

score 1 · Answer 2 · answered Oct 09 '17 at 18:15

1

$$\cos(\pi+i)=\frac{e^{i(\pi+i)}+e^{-i(\pi+i)}}{2}=\frac{e^{i\pi-1}+e^{1-i\pi}}{2}=\frac{-\frac{1}{e}-e}{2}=-\frac{1+e^2}{2e}\approx-1.5430806348...$$

answered Oct 09 '17 at 18:15

Thorgott

11,682

score 0 · Answer 3 · answered Oct 09 '17 at 18:21

Try this: $$ \begin{align} \cos(\pi+i) &=\frac{e^{i(\pi+i)}+e^{-i(\pi+i)}}{2}\\ &=\frac{e^{i\pi}e^{-1}+e^{-i\pi}e^1}{2}\\ &=-\frac{e^{1}+e^{-1}}{2}\\ &=-\cosh(1) \end{align} $$

Alternatively,

$$ \begin{align} \cos(\pi+i) &=\cos(\pi) \cos(i)-\sin(\pi)\sin(i)\\ &=-\cos(i)\\ &=-\cosh(1) \end{align} $$

Writing cos(π+i) in Cartesian form?

3 Answers3