complex no (1 Viewer)

[Mill]

Just trying to work it now.

Assume true for n = k:
ie. |z^k| = |z|^k

Prove true for n = k + 1:
ie. RTP: |z^(k+1)| = |z|^(k+1)
RHS = |z|^(k+1)
= |z|^k . |z|
= |z^k| . |z| using assumption
= I'm not sure where to go from here.

If i use rcis@ here, then it's simple, because you can simplify with de Moivre's theorem. The only thing I'm hung up on is that I don't think the question wants you to use mod-arg or cartesian form but rather just keep it as |z|, but I can't see another way to do it.

 

[Grey Council]

umm, what the hell. this seems a bit too easy. I think i'm doing something wrong:
Assume true for n = k:
ie. |z^k| = |z|^k

Prove true for n = k + 1:
ie. RTP: |z^(k+1)| = |z|^(k+1)

RHS = |z|^(k+1)
= |z|^k . |z|

LHS:
|z^(k+1)| = | z . z^k |
= |z| . |z^k|
= |z| . |z|^k
= RHS

qed

did i do anything wrong?

 

[shkspeare]

yea it does seem too easy.. lol

dumb q... just as dumb as those conjugate theorem proving crap

unless theres some other way to do it...

 

[Mill]

| z . z^k |
= |z| . |z^k|

Yeah that would work assuming we can use that complex law, and looking at it now, I imagine we can. I wasn't sure which ones we could and couldn't use - the question seems a bit finnicky.

 

[Grey Council]

ssuuurrreee! excuses, excuses, excuses.
so much for prior tution.

Heres an idea, everyone hire me. I'll tutor for $40 an hour. Its the price of quality. heheh

just joking Mill

anyway, as i said, i thought i was doing something wrong. If you can use the fact that I used, its easy. I was thinking the EXACT same thing as you, which complex laws are we allowed to use?

but I think generally the multiplication laws of moduli are taught before the power laws.

 

[martin310015]

..........isn't as hard as i thought it was

 

[Grey Council]

heheh