# Math Help - Analytic funtions

1. ## Analytic funtions

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

2. ## 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 $z = e^{a+bi} = e^a(\cos (b) + i \sin (b))$

3. ## Re: Analytic funtions

Originally Posted by ebaines
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?

4. ## Re: Analytic funtions

Q2 and Q3 - good!

Yes: $e^{\ln(e^z)} = z$

5. ## Re: Analytic funtions

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

6. ## Re: Analytic funtions

Originally Posted by Brennox
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?

7. ## Re: Analytic funtions

Originally Posted by ebaines
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

8. ## 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

9. ## 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:

$f'(z) = \frac {\delta u}{\delta x} + i \frac {\delta v }{\delta x} = \frac {\delta v}{\delta y} - i \frac {\delta u }{\delta y}$

10. ## Re: Analytic funtions

Originally Posted by Brennox
So im pretty sure i got part 1. 3 is the magnitude and its to the power of (5/6)(pi)i
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 $a=?~\&~b=?$

11. ## 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)"

12. ## Re: Analytic funtions

Originally Posted by Brennox
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 $u_x=v_y~\&~v_x=-u_y$.