Complex number question (1 Viewer)

[EvoRevolution]

Difficult:

Suppose θ, Φ doesnt = pie/2(2k+1) where k is an integer, use the fact that
z = (1+z)/(1+z^-1) to:

(a) find the real and imaginery parts of (1+cos2θ+isin2θ)/(1+cos2θ-isin2θ).

(b) show that if n is a positive integer then

[(1+sinΦ+isinΦ)/(1+sinΦ-icosΦ)]^n = cosn(pie/2-Φ)+isin n(pie/2-Φ).

 

[vds700]

 

[jet]

For <br> you just do //

 

[vds700]

For
you just do //

huh?

how do i type so it doesnt come up with br and ><?

 

[jet]

Sorry, the break i typed actually worked. Pardon, the residential was extremely tiring.
To do it without the tag showing, just use "//" [without the quotes obviously.]

 

[vds700]

Sorry, the break i typed actually worked. Pardon, the residential was extremely tiring.
To do it without the tag showing, just use "//" [without the quotes obviously.]

ok thanks bro

EDIT: doesnt seem to work either

 

[azureus88]

i think he means type // instead of that "br" crap

 

[jet]

Yeh. We need a latex guide. I might work on one and see what I come up with.

EDIT: Sorry. So tired. Its actually "\\"

 

[Trebla]

You're meant to connect the codes up into one. So when you type '\\', type the next line immediately after it, do not press enter.

 

[vds700]

You're meant to connect the codes up into one. So when you type '\\', type the next line immediately after it, do not press enter.

haha thanks, got it now

 

[Templar]

Yeh. We need a latex guide. I might work on one and see what I come up with.

EDIT: Sorry. So tired. Its actually "\\"

There are so many good guides online. Do a google search and you'll find a lot of them. Perhaps the mods should link a few to a sticky.

 

[Trebla]

(b) show that if n is a positive integer then

[(1+sinΦ+isinΦ)/(1+sinΦ-icosΦ)]^n = cosn(pie/2-Φ)+isin n(pie/2-Φ)

I think there is a mistake is should be show:
[(1+sinΦ+icosΦ)/(1+sinΦ-icosΦ)]^n = cosn(pie/2-Φ)+isin n(pie/2-Φ)

 

[gurmies]

^ I found the same, but wasn't sure - so kept trying with the sin's there -_-