Results 1 to 12 of 12
Like Tree3Thanks
  • 1 Post By ebaines
  • 1 Post By ebaines
  • 1 Post By ebaines

Math Help - Analytic funtions

  1. #1
    Junior Member
    Joined
    Mar 2012
    From
    USA
    Posts
    53

    Analytic funtions

    So im pretty sure i got part 1. 3 is the magnitude and its to the power of (5/6)(pi)i

    Please help with the following questions cheers.
    Attached Thumbnails Attached Thumbnails Analytic funtions-mathq11.png  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor ebaines's Avatar
    Joined
    Jun 2008
    From
    Illinois
    Posts
    1,103
    Thanks
    321

    Re: Analytic funtions

    Hint for 2 and 3:

     e^{ix} = \cos (x) + i\sin (x)

    So:

     e^{(a+bi)} = e^a e^{bi} = e^a( \cos (b) + i\sin (b))

    Hint for 4 and 5:

    If  \log(z) = a + bi, then  z = e^{a+bi} = e^a(\cos (b) + i \sin (b))
    Last edited by ebaines; May 9th 2013 at 07:43 AM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Mar 2012
    From
    USA
    Posts
    53

    Re: Analytic funtions

    Quote Originally Posted by ebaines View Post
    Hint for 2 and 3:

     e^{ix} = \cos (x) + i\sin (x)

    So:

     e^{(a+bi)} = e^a e^{bi} = e^a( \cos (b) + i\sin (b))

    Hint for 4 and 5:

    If  \log(z) = a + bi, then  z = e^{a+bi} = e^a(\cos (b) + i \sin (b))
    Q2 -9(3^0.5)/2

    Q3 9/2

    Q4 -3(3^0.5)/2

    im not sure with logs, is the log of "e^(a+bi)", just a + bi?
    Last edited by Brennox; May 9th 2013 at 09:36 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor ebaines's Avatar
    Joined
    Jun 2008
    From
    Illinois
    Posts
    1,103
    Thanks
    321

    Re: Analytic funtions

    Q2 and Q3 - good!

    Yes:  e^{\ln(e^z)} = z
    Thanks from Brennox
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Mar 2012
    From
    USA
    Posts
    53

    Re: Analytic funtions

    Thanks heaps, can you please help me with this Q.

    i can get part 1 but am really struggling with part 2. Do all values have to be <pi?

    heres link. http://oi41.tinypic.com/30mbqqe.jpg
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor ebaines's Avatar
    Joined
    Jun 2008
    From
    Illinois
    Posts
    1,103
    Thanks
    321

    Re: Analytic funtions

    Quote Originally Posted by Brennox View Post
    Do all values have to be <pi?
    No - they range between 0 and 2 pi.

    From de Moivre's formula, z^n = (R cis \theta )^n = R^n cis(n \theta), the angle  \theta of one solution will be 1/n the angle of z. Note that I like to use the notation "cis theta" (pronounced "kiss theta") as shorthand for  \cos \theta + i \sin \theta. But since cosine and sine functons have a period of  2 \pi, other solutions are available that are "evenly spaced" about the origin separated by the angle  \frac {2 \pi}{n}. In this way there are always n solutions to the nth root of z.

    Let me give an example: If z = 16i, which is equal to 16e^{ \frac {\pi}2 i} = 16 cis (\frac {\pi} 2), there are four 4th roots. The magnitude of z is 16, so the magnitude of the 4th roots are all  \sqrt[4] {16} = 2. The angle of the first solution is at angle  \theta = \pi /8, and the rest are at \pi/8 + k(2\pi/4), where k = 0, 1, 2, and 3. Thus the four 4th roots of 16i are:  2cis (\pi/8), \ 2 cis(5 \pi/8),\ 2 cis (9 \pi/8), \ 2 cis (13 \pi/8).

    Now, can you apply this thinking to the problem at hand?
    Last edited by ebaines; May 9th 2013 at 10:52 AM.
    Thanks from Brennox
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Mar 2012
    From
    USA
    Posts
    53

    Re: Analytic funtions

    Quote Originally Posted by ebaines View Post
    No - they range between 0 and 2 pi.

    From de Moivre's formula, z^n = (R cis \theta )^n = R^n cis(n \theta), the angle  \theta of one solution will be 1/n the angle of z. Note that I like to use the notation "cis theta" (pronounced "kiss theta") as shorthand for  \cos \theta + i \sin \theta. But since cosine and sine functons have a period of  2 \pi, other solutions are available that are "evenly spaced" about the origin separated by the angle  \frac {2 \pi}{n}. In this way there are always n solutions to the nth root of z.

    Let me give an example: If z = 16i, which is equal to 16e^{ \frac {\pi}2 i} = 16 cis (\frac {\pi} 2), there are four 4th roots. The magnitude of z is 16, so the magnitude of the 4th roots are all  \sqrt[4] {16} = 2. The angle of the first solution is at angle  \theta = \pi /8, and the rest are at \pi/8 + k(2\pi/4), where k = 0, 1, 2, and 3. Thus the four 4th roots of 16i are:  2cis (\pi/8), \ 2 cis(5 \pi/8),\ 2 cis (9 \pi/8), \ 2 cis (13 \pi/8).

    Now, can you apply this thinking to the problem at hand?
    thanks alot got it right
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Mar 2012
    From
    USA
    Posts
    53

    Re: Analytic funtions

    http://oi43.tinypic.com/34hz2ae.jpg

    for part 1 is...

    v = e^(5x)cos(5y)

    and how do i right an expression for f'(z) when there are x and y
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor ebaines's Avatar
    Joined
    Jun 2008
    From
    Illinois
    Posts
    1,103
    Thanks
    321

    Re: Analytic funtions

    Close - check your signs. I get  \frac {\delta u}{\delta x} = 5 e^{5x} \sin(5y) = \frac {\delta v}{\delta y}, so:

     v = \int 5 e^{5x} \sin(5y) \delta y = -e^{5x} \cos(5y)

    As for the derivative of a complex function z = u(x,y) + i v(x,y), it's:

    f'(z) = \frac {\delta u}{\delta x} + i \frac {\delta v }{\delta x} = \frac {\delta v}{\delta y} - i \frac {\delta u }{\delta y}
    Thanks from Brennox
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,705
    Thanks
    1637
    Awards
    1

    Re: Analytic funtions

    Quote Originally Posted by Brennox View Post
    So im pretty sure i got part 1. 3 is the magnitude and its to the power of (5/6)(pi)i
    Please help with the following questions cheers.
    I am greatly concerned about the author of these qyestions.
    Numbers 1,2, & 3 seem to make sense. But not # 4 or 5.

    By convention \log(z)=\log(|z|)+i\text{arg}(z). Thus \log \left( {\frac{{ - 3\sqrt 3 }}{2} + i\frac{3}{2}} \right) = \log (3) + i\left( {\frac{{5\pi }}{6}} \right)

    With that definition e^{\log(z)}=z and e^{e^{\log(z)}}=e^z\ne z.

    Thus if \log(z)=a+bi then a=?~\&~b=?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Junior Member
    Joined
    Mar 2012
    From
    USA
    Posts
    53

    Re: Analytic funtions

    thanks heaps, i dont understand why i got this Q wrong tho.

    http://oi43.tinypic.com/d4suq.jpg

    du/dx = 2x
    dv/dy = 2x

    du/dy = -2y + 5
    dv/dx = -2y + 5

    Did i do the wrong input for part 2? "x^2-y^2+5y+9+i(-2xy+5x)"
    Follow Math Help Forum on Facebook and Google+

  12. #12
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,705
    Thanks
    1637
    Awards
    1

    Re: Analytic funtions

    Quote Originally Posted by Brennox View Post
    thanks heaps, i dont understand why i got this Q wrong tho.
    http://oi43.tinypic.com/d4suq.jpg
    You have it wrong because you have part #1 wrong. The answer is b not c.

    The final answer is u(x,y)+iv(x,y)=(x^2-y^2+5y+9)+i(2xy-5x).

    Remember, you need to have u_x=v_y~\&~v_x=-u_y.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: March 5th 2012, 04:50 PM
  2. Inverse funtions
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: March 7th 2009, 10:00 PM
  3. types of funtions
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: September 15th 2008, 05:23 PM
  4. inverse funtions
    Posted in the Calculus Forum
    Replies: 1
    Last Post: March 10th 2008, 04:37 PM
  5. sum of two lipschitz funtions
    Posted in the Calculus Forum
    Replies: 2
    Last Post: November 13th 2007, 02:24 PM

Search Tags


/mathhelpforum @mathhelpforum