Turn on thread page Beta
    • Thread Starter
    Offline

    13
    ReputationRep:
    Hi, I've got to evaluate (root3 - i)^4 divided by (-1 + i)^3, I've so far gotten to (-2 - 2root3) + (2 - 2root3)i, but I have to convert it to exponential form, and the question is non-calculator.
    Offline

    18
    ReputationRep:
    If your answer is supposed to be in the form r e^{i\theta}, then you need to convert \sqrt{3} - i and -1 + i into this form and then use DeMoivre. It's not difficult - they are all "standard" angles.
    Offline

    21
    ReputationRep:
    (Original post by jordanwu)
    Hi, I've got to evaluate (root3 - i)^4 divided by (-1 + i)^3, I've so far gotten to (-2 - 2root3) + (2 - 2root3)i, but I have to convert it to exponential form, and the question is non-calculator.
    Draw a diagram and convert each complex number to the form re^{i\theta}. That makes applying the powers, and dividing, a lot easier.
    • Thread Starter
    Offline

    13
    ReputationRep:
    (Original post by DFranklin)
    If your answer is supposed to be in the form r e^{i\theta}, then you need to convert \sqrt{3} - i and -1 + i into this form and then use DeMoivre. It's not difficult - they are all "standard" angles.
    Ok, but I'm not sure if I've got the right exponential forms: 2e^i11/6pi and root2e^i3/4pi?
    Offline

    15
    ReputationRep:
    (Original post by jordanwu)
    Ok, but I'm not sure if I've got the right exponential forms: 2e^i11/6pi and root2e^i3/4pi?
    Yeah those are correct
    • Thread Starter
    Offline

    13
    ReputationRep:
    (Original post by DFranklin)
    If your answer is supposed to be in the form r e^{i\theta}, then you need to convert \sqrt{3} - i and -1 + i into this form and then use DeMoivre. It's not difficult - they are all "standard" angles.
    Not too sure about how to use that on the exponential form
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by jordanwu)
    Not too sure about how to use that on the exponential form
    Well, if \sqrt{3}-i=2e^{-\frac{1}{6}\pi i} as you found, then (\sqrt{3}-i)^4=....?
    • Thread Starter
    Offline

    13
    ReputationRep:
    (Original post by RDKGames)
    Well, if \sqrt{3}-i=2e^{-\frac{1}{6}\pi i} as you found, then (\sqrt{3}-i)^4=....?
    Oh it's supposed to be -11/6pi? I've got 11/6pi
    Offline

    20
    ReputationRep:
    (Original post by jordanwu)
    Oh it's supposed to be -11/6pi? I've got 11/6pi
    if you have w = re

    then wk = rkeikΘ
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by jordanwu)
    Oh it's supposed to be -11/6pi? I've got 11/6pi
    e^{\frac{11}{6}\pi i}=e^{-\frac{1}{6}\pi i}
    • Thread Starter
    Offline

    13
    ReputationRep:
    (Original post by RDKGames)
    e^{\frac{11}{6}\pi i}=e^{-\frac{1}{6}\pi i}
    Anywhere I can look that rule up? I've definitely forgotten it. So I have to do 2^4 and (e^-1/6pi*i)^4? And same with the exponential on the denominator?
    Offline

    21
    ReputationRep:
    (Original post by jordanwu)
    Anywhere I can look that rule up? I've definitely forgotten it. So I have to do 2^4 and (e^-1/6pi*i)^4? And same with the exponential on the denominator?
    You can remove (or add) any number of integer multiples of 2\pi, as it's removing complete revolutions from the angle.
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by jordanwu)
    Anywhere I can look that rule up? I've definitely forgotten it. So I have to do 2^4 and (e^-1/6pi*i)^4? And same with the exponential on the denominator?
    An exponential form complex number e^{i \theta} repeats itself every 2\pi so e^{i \theta}=e^{i (2\pi k + \theta)} for n \in \mathbb{Z}

    Pick \theta= \frac{11}{6} \pi and k=-1 and you arrive at the result I've shown. However, it is much more intuitive to mark down the complex number \sqrt{3}-i on the complex plane, draw a line to it (its modulus), then notice how your angle is simply measured anticlockwise from \mathbb{R}_{>0} while my angle is just measuring backwards (clockwise, hence the negative) to get my solution.


    Anywho, yes you simply exponentiate these numbers to 4, then repeat similarly for the other complex number. Then you're simply in a position where you have \displaystyle \frac{re^{i \theta}}{Re^{i \phi}} which should be straight-forward to evaluate.
    • Thread Starter
    Offline

    13
    ReputationRep:
    (Original post by RDKGames)
    An exponential form complex number e^{i \theta} repeats itself every 2\pi so e^{i \theta}=e^{i (2\pi k + \theta)} for n \in \mathbb{Z}

    Pick \theta= \frac{11}{6} \pi and k=-1 and you arrive at the result I've shown. However, it is much more intuitive to mark down the complex number \sqrt{3}-i on the complex plane, draw a line to it (its modulus), then notice how your angle is simply measured anticlockwise from \mathbb{R}_{>0} while my angle is just measuring backwards (clockwise, hence the negative) to get my solution.


    Anywho, yes you simply exponentiate these numbers to 4, then repeat similarly for the other complex number. Then you're simply in a position where you have \displaystyle \frac{re^{i \theta}}{Re^{i \phi}} which should be straight-forward to evaluate.
    I'm not sure how to do (e^(11/6pi)i)^4
    Offline

    18
    ReputationRep:
    (Original post by jordanwu)
    I'm not sure how to do (e^(11/6pi)i)^4
    (e^a)^b = e^{ab}...
    • Thread Starter
    Offline

    13
    ReputationRep:
    (Original post by DFranklin)
    (e^a)^b = e^{ab}...
    e^(22/3*pi^4*i^4)?
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by jordanwu)
    e^(22/3*pi^4*i^4)?
    Ah dear, someone's forgot a GCSE rule.

    You have z=(\sqrt{3}-i)^4=(2e^{\frac{11}{6}\pi i})^4

    So... since (e^a)^b=e^{ab}, you simply have z=2^4e^{\frac{44}{6}\pi i}
    Offline

    18
    ReputationRep:
    (Original post by jordanwu)
    e^(22/3*pi^4*i^4)?
    No. Compare (e^{(11/6\pi)i})^4 with the form (e^a)^b.

    What is a here?
    What is b here?

    So, then what is ab?

    So what is e^{ab}?
    • Thread Starter
    Offline

    13
    ReputationRep:
    (Original post by RDKGames)
    Ah dear, someone's forgot a GCSE rule.

    You have z=(\sqrt{3}-i)^4=(e^{\frac{11}{6}\pi i})^4

    So... since (e^a)^b=e^{ab}, you simply have z=e^{\frac{44}{6}\pi i}
    Wow, what's wrong with me... don't even know how I got that. Ok, 16e^44/6pi*i / 2root2e^9/4pi*i?
    • Community Assistant
    Offline

    20
    ReputationRep:
    Community Assistant
    (Original post by jordanwu)
    Wow, what's wrong with me... don't even know how I got that. Ok, 16e^44/6pi*i / 2root2e^9/4pi*i?
    Yeah
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: November 15, 2017
Poll
Do you think parents should charge rent?
Useful resources

Make your revision easier

Maths

Maths Forum posting guidelines

Not sure where to post? Read the updated guidelines here

Equations

How to use LaTex

Writing equations the easy way

Student revising

Study habits of A* students

Top tips from students who have already aced their exams

Study Planner

Create your own Study Planner

Never miss a deadline again

Polling station sign

Thinking about a maths degree?

Chat with other maths applicants

Can you help? Study help unanswered threads

Groups associated with this forum:

View associated groups

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Write a reply...
Reply
Hide
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.