It is currently Wed Sep 20, 2017 7:19 am

 4 posts • Page 1 of 1
Author Message
Prometheus
 Post subject: Complex numbers question  |  Posted: Thu Jan 19, 2012 3:12 pm

Original Member

Joined: Sun Aug 07, 2011 8:58 am
Posts: 314

 Doing a remedial maths course and i just want to check this question.I need to state the real part $\displaystyle x$ and the imaginary part $\displaystyle y$ of:$\displaystyle Z=(-i)^9$I assume they are expecting me to do it by inspection given it's only worth 1 mark, in which case i make it$\displaystyle x=0$ and $\displaystyle y=-1$ since it's already in Cartesian form.But i wanted to prove it to myself to make sure i understand the concept not just a mechanical application. So i changed it to polar exponential form in order to expand the brackets and i just want to make sure the workings are correct:$\displaystyle r=\sqrt0^2+\sqrt-1^2=1$$\displaystyle \theta=tan^{-1} \frac{-1}{0}= -\frac{\pi}{2}Is the latter correct, even though it is technically undefined, when thinking of the unit circle it equates to \displaystyle \frac{\pi}{2}?If so, then \displaystyle (e^{-i\frac{\pi}{2}})^9=e^-{i\frac{9\pi}{2}}and putting back into Cartesian from: \displaystyle cos(-\frac{9\pi}{2})+isin(-\frac{9\pi}{2})Which recovers \displaystyle Z=0-i1ThanksEdit: - not + DrRocket  Post subject: Re: Complex numbers question | Posted: Sun Jan 22, 2012 3:43 am Original Member Joined: Fri Aug 05, 2011 2:22 am Posts: 477  Prometheus wrote:Doing a remedial maths course and i just want to check this question.I need to state the real part \displaystyle x and the imaginary part \displaystyle y of: \displaystyle Z=(-i)^9I assume they are expecting me to do it by inspection given it's only worth 1 mark, in which case i make it \displaystyle x=0 and \displaystyle y=-1 since it's already in Cartesian form.But i wanted to prove it to myself to make sure i understand the concept not just a mechanical application. So i changed it to polar exponential form in order to expand the brackets and i just want to make sure the workings are correct: \displaystyle r=\sqrt0^2+\sqrt-1^2=1$$ \displaystyle \theta=tan^{-1} \frac{-1}{0}= -\frac{\pi}{2}$Is the latter correct, even though it is technically undefined, when thinking of the unit circle it equates to $\displaystyle \frac{\pi}{2}$?If so, then $\displaystyle (e^{-i\frac{\pi}{2}})^9=e^-{i\frac{9\pi}{2}}$and putting back into Cartesian from:$\displaystyle cos(-\frac{9\pi}{2})+isin(-\frac{9\pi}{2})$Which recovers $\displaystyle Z=0-i1$ThanksEdit: - not +That is correct, but an awfully hard way to do it.It is a bit easier to note that $\displaystyle -i^n = -1^ni^n$ and hence $\displaystyle -i^9 = -1^9i^9= -1 i^8i = -1(i^2)^4i = -1(-1)^4i = -i$So the real part is 0 and the imaginary part is -1. _________________gone
Prometheus
 Post subject: Re: Complex numbers question  |  Posted: Sun Jan 22, 2012 1:42 pm

Original Member

Joined: Sun Aug 07, 2011 8:58 am
Posts: 314

 That's a lot easier, thanks. Love how there can be many ways to get the same answer, though my tendency is to pick the longest. Hopefully that'll change with understanding.
DrRocket
 Post subject: Re: Complex numbers question  |  Posted: Mon Jan 23, 2012 12:24 am
Original Member

Joined: Fri Aug 05, 2011 2:22 am
Posts: 477

 Prometheus wrote:That's a lot easier, thanks. Love how there can be many ways to get the same answer, though my tendency is to pick the longest. Hopefully that'll change with understanding.There is no single right way to solve a mathematical problem. With experience you wil be able to see the shorter and more elegant approaches to a given problem. Usually (not always, but usually) the shortest and most elegant approach is also the most clear. _________________gone
 Display posts from previous: All posts1 day7 days2 weeks1 month3 months6 months1 year Sort by AuthorPost timeSubject AscendingDescending
 4 posts • Page 1 of 1

Who is online
 Users browsing this forum: No registered users and 0 guests You cannot post new topics in this forumYou cannot reply to topics in this forumYou cannot edit your posts in this forumYou cannot delete your posts in this forum

 Delete all board cookies | The team | All times are UTC

 This free forum is proudly hosted by ProphpBB | phpBB software | Report Abuse | Privacy