Page 1 of 2 12 LastLast
Results 1 to 15 of 17

Math Help - Complex Numbers raised to the power?

  1. #1
    Newbie
    Joined
    Feb 2013
    From
    islamabad
    Posts
    23

    Complex Numbers raised to the power?

    If v = z(1)^6* z(2)^−4
    determine |v| and angle(v). Also, obtain v in cartesian form.

    where z1= 3 +i
    z2 = -5 + 5i

    I don't know how to raise a complex power to a power. I have searched all over but nothing online makes sense.
    Please help me. Thanks!

    So far I have done |(√10)^6 * (5√2)^-4| = .4

    If that is correct please explain how to solve for angles.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Sep 2012
    From
    Australia
    Posts
    3,807
    Thanks
    660

    Re: Complex Numbers raised to the power?

    Hey kmalik001.

    If you have a complex number z = r*(cos(theta) + i*sin(theta)) then z^n = r^n * (cos(n*theta) + i*sin(n*theta)).
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,790
    Thanks
    1687
    Awards
    1

    Re: Complex Numbers raised to the power?

    Quote Originally Posted by kmalik001 View Post
    If v = z(1)^6* z(2)^−4
    determine |v| and angle(v). Also, obtain v in cartesian form.
    where z1= 3 +i
    z2 = -5 + 5i
    If that is correct please explain how to solve for angles.

    Frankly, I do not know what that says.

    Is it (3+i)^6\cdot(-5+5i)^{-4}~?

    It is foolish to try to guess as to what you really meant to post.
    So please answer this.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Feb 2013
    From
    islamabad
    Posts
    23

    Re: Complex Numbers raised to the power?

    Okay well for the first z1^6 I got 1000cos(110.58) + isin(-270)

    So 2 questions:

    Should the angles be in radians?

    How am i going to multiply this answer to when I get z2^-4. Because I can not change it to re^jtheta form because the angles in cos and sin are different
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Feb 2013
    From
    islamabad
    Posts
    23

    Re: Complex Numbers raised to the power?

    like one angle is 110.58 and the other is -270
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Feb 2013
    From
    islamabad
    Posts
    23

    Re: Complex Numbers raised to the power?

    Yes that is what is says Plato
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,790
    Thanks
    1687
    Awards
    1

    Re: Complex Numbers raised to the power?

    Quote Originally Posted by kmalik001 View Post
    like one angle is 110.58 and the other is -270
    Your mistake is to use degrees.

    Let \theta=\arctan\left(\frac{1}{3}\right).

    Now z_1=\sqrt{10}\exp(\theta i)~\&~z_2=5\sqrt{2}\exp\left(\frac{-3\pi i}{4}\right)

    (z_1)^6=10^3\exp(6\theta i) and (z_2)^{-4}=5^{-4}(2^{-2})\exp\left({-3\pi i}\right).
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Newbie
    Joined
    Feb 2013
    From
    islamabad
    Posts
    23

    Re: Complex Numbers raised to the power?

    Just one question are you sure you did your second angle -3pi correct?

    Because (-3pi/4) ^-4 is equal to 3.16*pi which isn't equal to 3*pi

    I just checked on the calculator but I want to double check with you
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Feb 2013
    From
    islamabad
    Posts
    23

    Re: Complex Numbers raised to the power?

    The second part of this question asks me to solve for z1^3000. I tried the same method but when I type in 3000 power in calculator it doesn't allow me is there another method I could use. Taylor's method or binomial theorem I always get confused solving
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,790
    Thanks
    1687
    Awards
    1

    Re: Complex Numbers raised to the power?

    Quote Originally Posted by kmalik001 View Post
    The second part of this question asks me to solve for z1^3000. I tried the same method but when I type in 3000 power in calculator it doesn't allow me is there another method I could use. Taylor's method or binomial theorem I always get confused solving

    Are you very sure that z_1 is not given as \sqrt{3}+i~?

    That is the only way (z_1)^{3000} makes sense as a question.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Newbie
    Joined
    Feb 2013
    From
    islamabad
    Posts
    23

    Re: Complex Numbers raised to the power?

    This is the exact question.

    If w = z1^(3000) determine angle of w in the range [0, 2pi). Your answer should be correct to four decimal
    places.
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,790
    Thanks
    1687
    Awards
    1

    Re: Complex Numbers raised to the power?

    Quote Originally Posted by kmalik001 View Post
    This is the exact question.
    If w = z1^(3000) determine angle of w in the range [0, 2pi). Your answer should be correct to four decimal places.

    That was NOT what I asked you could it be that z_1=\sqrt{3}+i~?
    Not 3+i but \sqrt{3}+i.
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Newbie
    Joined
    Feb 2013
    From
    islamabad
    Posts
    23

    Re: Complex Numbers raised to the power?

    Yes, z1 = 3 + i is the question
    Follow Math Help Forum on Facebook and Google+

  14. #14
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,790
    Thanks
    1687
    Awards
    1

    Re: Complex Numbers raised to the power?

    Quote Originally Posted by kmalik001 View Post
    This is the exact question.

    If w = z1^(3000) determine angle of w in the range [0, 2pi). Your answer should be correct to four decimal places.

    Well then, \text{arg}\left( {3 + i} \right)^{3000}=\bmod \left( {3000\arctan \left( {\frac{1}{3}} \right),2\pi } \right) = {\text{3}}{\text{.9243}}
    Follow Math Help Forum on Facebook and Google+

  15. #15
    Newbie
    Joined
    Feb 2013
    From
    islamabad
    Posts
    23

    Re: Complex Numbers raised to the power?

    I understand the 3000* arctan(1/3)

    but what did you do after that to get 3.92 like what does the ,2pi mean?
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Power and root of complex numbers?
    Posted in the Pre-Calculus Forum
    Replies: 8
    Last Post: May 2nd 2011, 02:52 PM
  2. Integer Raised To Complex Number Power
    Posted in the Pre-Calculus Forum
    Replies: 13
    Last Post: April 23rd 2010, 07:55 PM
  3. Raising Complex Numbers to nth power.
    Posted in the Calculus Forum
    Replies: 5
    Last Post: January 29th 2010, 10:44 AM
  4. Log raised to a power
    Posted in the Algebra Forum
    Replies: 5
    Last Post: January 5th 2010, 08:09 AM
  5. Complex Numbers and power series
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: January 4th 2010, 03:10 AM

Search Tags


/mathhelpforum @mathhelpforum