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Thread: Analytic funtions

  1. #1
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    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  
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  2. #2
    MHF Contributor ebaines's Avatar
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    Re: Analytic funtions

    Hint for 2 and 3:

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

    So:

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

    Hint for 4 and 5:

    If $\displaystyle \log(z) = a + bi$, then $\displaystyle z = e^{a+bi} = e^a(\cos (b) + i \sin (b))$
    Last edited by ebaines; May 9th 2013 at 07:43 AM.
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  3. #3
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    Re: Analytic funtions

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

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

    So:

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

    Hint for 4 and 5:

    If $\displaystyle \log(z) = a + bi$, then $\displaystyle z = e^{a+bi} = e^a(\cos (b) + i \sin (b))$
    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.
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  4. #4
    MHF Contributor ebaines's Avatar
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    Re: Analytic funtions

    Q2 and Q3 - good!

    Yes: $\displaystyle e^{\ln(e^z)} = z$
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    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
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  6. #6
    MHF Contributor ebaines's Avatar
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    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, $\displaystyle z^n = (R cis \theta )^n = R^n cis(n \theta)$, the angle $\displaystyle \theta$ 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 $\displaystyle \cos \theta + i \sin \theta$. But since cosine and sine functons have a period of $\displaystyle 2 \pi$, other solutions are available that are "evenly spaced" about the origin separated by the angle $\displaystyle \frac {2 \pi}{n}$. 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 $\displaystyle 16e^{ \frac {\pi}2 i} = 16 cis (\frac {\pi} 2)$, there are four 4th roots. The magnitude of z is 16, so the magnitude of the 4th roots are all $\displaystyle \sqrt[4] {16} = 2$. The angle of the first solution is at angle $\displaystyle \theta = \pi /8$, and the rest are at $\displaystyle \pi/8 + k(2\pi/4)$, where k = 0, 1, 2, and 3. Thus the four 4th roots of 16i are:$\displaystyle 2cis (\pi/8), \ 2 cis(5 \pi/8),\ 2 cis (9 \pi/8), \ 2 cis (13 \pi/8)$.

    Now, can you apply this thinking to the problem at hand?
    Last edited by ebaines; May 9th 2013 at 10:52 AM.
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  7. #7
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    Re: Analytic funtions

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

    From de Moivre's formula, $\displaystyle z^n = (R cis \theta )^n = R^n cis(n \theta)$, the angle $\displaystyle \theta$ 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 $\displaystyle \cos \theta + i \sin \theta$. But since cosine and sine functons have a period of $\displaystyle 2 \pi$, other solutions are available that are "evenly spaced" about the origin separated by the angle $\displaystyle \frac {2 \pi}{n}$. 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 $\displaystyle 16e^{ \frac {\pi}2 i} = 16 cis (\frac {\pi} 2)$, there are four 4th roots. The magnitude of z is 16, so the magnitude of the 4th roots are all $\displaystyle \sqrt[4] {16} = 2$. The angle of the first solution is at angle $\displaystyle \theta = \pi /8$, and the rest are at $\displaystyle \pi/8 + k(2\pi/4)$, where k = 0, 1, 2, and 3. Thus the four 4th roots of 16i are:$\displaystyle 2cis (\pi/8), \ 2 cis(5 \pi/8),\ 2 cis (9 \pi/8), \ 2 cis (13 \pi/8)$.

    Now, can you apply this thinking to the problem at hand?
    thanks alot got it right
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    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
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    Re: Analytic funtions

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

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

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

    $\displaystyle f'(z) = \frac {\delta u}{\delta x} + i \frac {\delta v }{\delta x} = \frac {\delta v}{\delta y} - i \frac {\delta u }{\delta y} $
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    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 $\displaystyle \log(z)=\log(|z|)+i\text{arg}(z)$. Thus $\displaystyle \log \left( {\frac{{ - 3\sqrt 3 }}{2} + i\frac{3}{2}} \right) = \log (3) + i\left( {\frac{{5\pi }}{6}} \right)$

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

    Thus if $\displaystyle \log(z)=a+bi$ then $\displaystyle a=?~\&~b=?$
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    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)"
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  12. #12
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    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 $\displaystyle u(x,y)+iv(x,y)=(x^2-y^2+5y+9)+i(2xy-5x)$.

    Remember, you need to have $\displaystyle u_x=v_y~\&~v_x=-u_y$.
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