Cos(-i) in the form a+bi

leonmate

Show cos(-i) in the form a+bi

I've been working on a worksheet of this kind of question but up until now I have been using De Moiuvre's theorem

e.g

(1 + i)^8

First, i set z = 1 + i

then the modulus of z = [itex]\sqrt{}2[/itex]

Next i need to find θ = arctan[itex]1/1[/itex] = [itex]∏/2[/itex][/quote]
arctan(1)= [itex]\pi/4[/itex], not [itex]\pi/2[/itex].

[quote]Then I can represent the equation in polar form,

z = [itex]\sqrt{}2[/itex] [cos([itex]∏/2[/itex]) + i sin([itex]∏/2[/itex])]

z^8 = [itex]\sqrt{}2[/itex]^8 [cos([itex]8∏/2[/itex]) + i sin([itex]8∏/2[/itex])]

z^8 = 16

(1 + i)^8 = 16 + 0i


Basically what I'm asking for is a method than would allow me to convert trigonemetric functions of complex numbers into the rectangular form a + bi

Thanks!

Simon Bridge

Boy that comes out really ugly!
FYI: The lower-case pi in the chart looks like the letter n. It's this one "π".
Or, better(!), use LaTeX.

for instance: [itex] z=\sqrt{2}\big [ \cos(\pi/2)+i\sin(\pi/2) \big ] [/itex]

Anyway - I'd use Euler's relation, expressing the trig function as exponentials.

jahaan

If you know the identity [itex]\cos x = \frac{e^{i x}+ e^{-ix}}{2}[/itex], you get:

[itex]\cos -i = \frac{e^{-i^2}+ e^{+i^2}}{2}=\frac{e^{1}+ e^{-1}}{2}=\cosh 1[/itex]

So it's a purely real number ;-)

leonmate

Haha, yeah i though pi looked huge, yours looks pretty :P

Would you mind elaborating a bit on that? Im not really sure how to swap the cos for an exponential, the only euler's relation im aware of is

e^iθ = [cos(θ) + i sin(θ)]

EDIT: just seen jahaan's comment, thank you!

Simon Bridge

No worries - I was going to make you work it out for yourself since we are not supposed to do your work for you.
Jahaan spoiled my cunning plan - need more weasels!

Euler relation -> trig.
Also work out [itex]e^{-i\theta}[/itex] as per the Euler relation.
Gives you the other half of a pair of simultaneous equations that you can solve for sin or cos in terms of complex exponentials.

To see how we get the math to look so pretty, just click "quote" at the bottom of a post and compare the stuff in the "tex" or "itex" tags with what you see.

jahaan

Ok, I'm sorry I'm new to this forum (it was just my 6th post), but i'll stick to hints next time!

Simon Bridge

Not a problem - but if you've ever wondered why the first few answers tend to seem oddly unhelpful... now you know :D

Bohrok

Start with [itex]e^{i\theta} = \cos\theta + i\sin\theta[/itex] and [itex]e^{-i\theta} = \cos\theta - i\sin\theta[/itex]

Then you add the equations and get[tex]e^{i\theta} + e^{-i\theta} = 2\cos\theta \Longrightarrow \cos\theta = \frac{e^{i\theta} + e^{-i\theta}}{2}[/tex]

Simon Bridge

It's so cool - nobody ever gets to do their own math around here ;)

leonmate

That's really really helpful thanks for clearing this up. Revising for a maths retake at the moment, just done my first year of a physics degree. Think i will be visiting this forum a lot more!

jahaan

By the way, the Euler relation is a very quick and handy way to derive any trigonometric identities you might forget during an exam. Try it ;-)

NegativeDept

The Taylor-series representation of ##e^z## is a very powerful trick for this sort of thing, especially when combined with the formulas other people mentioned.

##
\exp(z) = 1 + z + \tfrac{1}{2}z^2 + \tfrac{1}{3!}z^3 + \cdots + \tfrac{1}{k!}z^k + \cdots
##

This series converges for any real or complex ##z##. It can also be used to define trig functions and hyperbolic trig functions for real or complex numbers:

##\cosh(z)## = the even terms of ##\exp(z)## (including 1 as the 0th term)
##\sinh(z)## = the odd terms of ##\exp(z)##

##\cos(z)## = the even terms of ##\exp(\imath z)## (including 1 as the 0th term)
##\imath \sin(z)## = the odd terms of ##\exp(\imath z)##

For example, ##\cos(\imath)## = the even terms of ##\exp(\imath^2)##
##
= 1 + \tfrac{1}{2}(-1)^2 + \tfrac{1}{4!}(-1)^4 + \cdots
= 1 + \tfrac{1}{2} + \tfrac{1}{4!} + \cdots = \cosh(1)
##