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.

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- May 9th 2013, 06:58 AMBrennoxAnalytic 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. - May 9th 2013, 07:38 AMebainesRe: 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 $\displaystyle z = e^{a+bi} = e^a(\cos (b) + i \sin (b))$ - May 9th 2013, 09:17 AMBrennoxRe: Analytic funtions
- May 9th 2013, 09:27 AMebainesRe: Analytic funtions
Q2 and Q3 - good!

Yes: $\displaystyle e^{\ln(e^z)} = z$ - May 9th 2013, 10:05 AMBrennoxRe: 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 - May 9th 2013, 10:36 AMebainesRe: Analytic funtions
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? - May 9th 2013, 11:19 AMBrennoxRe: Analytic funtions
- May 9th 2013, 11:34 AMBrennoxRe: 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 - May 9th 2013, 12:08 PMebainesRe: 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} $ - May 9th 2013, 12:14 PMPlatoRe: Analytic funtions
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 $\displaystyle a=?~\&~b=?$ - May 9th 2013, 11:11 PMBrennoxRe: 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)" - May 10th 2013, 04:27 AMPlatoRe: Analytic funtions