The zeros of an analytic function (Complex Analysis)

**shadow_2145** · October 23rd 2008, 09:25 AM

Determine the number of zeros of $f$ in the first quadrant.

(1) $f(z)=z^4-3z^2+3$

(2) $f(z)=z^2+iz+2+i$

(3) $f(z) = z^7+6z^3+7$

I really can't follow along in the book of how they do this. I was wondering if there was a simpler way of figuring out these problems. Any help would be appreciated, thanks!

**shadow_2145** · October 23rd 2008, 01:26 PM

Here is the example they give me to follow:

Find the number of zeros of the function $f(z)=z^3-2z^2+4$ in the first quadrant

Solution: On the segment $0\leq x\leq R$ , $f(x) = x^3-2x^2+4$ is real and greater than 2. On the quartercircle $z=Re^{it}$ , $0\leq t\leq \frac{\pi}{2}$ ,

$f(Re^{it}) = R^3e^{3it}(1-\frac{2}{Re^{it}}+\frac{4}{R^3e^{3it}})=R^3e^{it}( 1+\zeta)$ ,

where $|\zeta|\leq \frac{6}{R} <\epsilon$ for $R$ large. Thus, arg $f(Re^{it})$ is approx. arg $(e^{3it}) = 3t$ for large $R$ , so arg $f(Re^{it})$ increases from 0 to about $\frac{3\pi}{2}$ as $t$ increases from 0 to $\frac{\pi}{2}$

On the segment $z=iy$ , $R\geq y\geq 0$ ,

$f(iy)=-iy^3+2y^2+4$

For $R\geq y > 0$ , this point lies in the fourth quadrant, since

Re $(f(iy))=4+2y^2 > 0$
Im $(f(iy))=-y^3 < 0$

Hence, as y decreases from R to 0, f(iy) lies in the fourth quadrant and moves toward the point w=4. COnsequently, as z traverses the contour, arg f(z) increases by exactly 2 $\pi$ , so $f(z)=z^3-2z^2+4$ has precisely one zero in the first quadrant.
--------------------------------------------------------------------

I can follow these steps up until where $|\zeta| \leq \frac{6}{R} < \epsilon$ .

Thanks

**shawsend** · October 23rd 2008, 02:52 PM

That's a nice analysis Shadow. Thanks. This is how I'd interpret it: We're of course using the Argument Principle: $\Delta_C \text{arg} f(z)=2\pi(N-P)$ .

In the case of the quarter circle $C_2=\left\{z=Re^{it},\; 0\leq t\leq \pi/2\right\}$ you have:

$<br /> f(Re^{it}) = R^3e^{3it}(1-\frac{2}{Re^{it}}+\frac{4}{R^3e^{3it}})=R^3e^{3it} (1+\zeta)<br />$

Now: $\left|\frac{2}{Re^{it}}+\frac{4}{R^3e^{3it}}\right |\leq \frac{2}{R}+\frac{4}{R^3}\leq \frac{2}{R}+\frac{4}{R}=\frac{6}{R}$ and so for large R,we have in parenthesis, one minus a very small number which approaches one in the limit so that we have $\Delta_{C_2}\text{arg}f(z)=\Delta_{C_2}\text{arg}R e^{3it}$ over the quarter circle which is just $3\pi/2$ .

On the leg going down the imaginary axis you noted:
$\text{Re} f(iy)=4+2y^2>0$
$\text{Im} f(iy)=-y^3<0$
So for very large $R$ , the argument of this point tends to $-\pi/2$ since it's in the 4th quad and the $-y^3$ term dominates for large y. As we go towards the origin, $f(0)=4$ so the change in argument is $0-(-\pi/2)=\pi/2$ .

Then the total change in argument is $2\pi$ which implies there is only one root in the first quadrant.

You might be interested to know that Riemann did a similar, albeit slightly more complicated, analysis of the zeta function around a square contour encircling the critical region and likewise using the argument principle, conjectured all the (non-trivial) zeros of zeta were on the critical line.

**shadow_2145** · October 23rd 2008, 03:07 PM

Thanks a lot! I'm going to try these on my own. Is there any chance you could tell me the answer to at least one of these so I can see if I'm going on the right track? Just the answer is fine, no ned to explain if you don't want to.

**shawsend** · October 23rd 2008, 03:19 PM

Originally Posted by shadow_2145

Thanks a lot! I'm going to try these on my own. Is there any chance you could tell me the answer to at least one of these so I can see if I'm going on the right track? Just the answer is fine, no ned to explain if you don't want to.

You mean the other two right? You just did the first one. I struggle with them too. As far as the answers well, got Mathematica?

Code:

In[69]:= N[Solve[z^2 + I*z + 2 + I == 0, z], 4] 
N[Solve[z^7 + 6*z^3 + 7 == 0, z], 4]


Out[69]= {{z -> -0.3257 + 1.0350 I}, {z -> 
   0.326 - 2.035 I}}

Out[70]= {{z -> -1.000}, {z -> -1.122 - 
    1.201 I]}, {z -> -1.122 + 1.201 I}, {z -> 
   0.4644 - 0.9559 I}, {z -> 
   0.4644 + 0.9559 I]}, {z -> 
   1.157 - 0.978 I}, {z -> 1.157 + 0.978 I}}

Yea, yea I know it's cheezy to do that. I ain't proud. It does provide a check of the work we do using the Argument Principle though.

**shadow_2145** · October 23rd 2008, 03:33 PM

Actually, the first one is different from the example I provided. What is this mathematica?! I don't understand it, lol.

**shadow_2145** · October 23rd 2008, 03:38 PM

Originally Posted by shawsend

You mean the other two right? You just did the first one. I struggle with them too. As far as the answers well, got Mathematica?

Code:

In[69]:= N[Solve[z^2 + I*z + 2 + I == 0, z], 4] 
N[Solve[z^7 + 6*z^3 + 7 == 0, z], 4]
 
 
Out[69]= {{z -> -0.3257 + 1.0350 I}, {z -> 
   0.326 - 2.035 I}}
 
Out[70]= {{z -> -1.000}, {z -> -1.122 - 
    1.201 I]}, {z -> -1.122 + 1.201 I}, {z -> 
   0.4644 - 0.9559 I}, {z -> 
   0.4644 + 0.9559 I]}, {z -> 
   1.157 - 0.978 I}, {z -> 1.157 + 0.978 I}}

Yea, yea I know it's cheezy to do that. I ain't proud. It does provide a check of the work we do using the Argument Principle though.

So [69] is 2 zeros and [70] is 7 zeros? :/

**shawsend** · October 23rd 2008, 03:43 PM

Originally Posted by shadow_2145

Actually, the first one is different from the example I provided. What is this mathematica?! I don't understand it, lol.

Ok, I didn't notice. That's good though. Gives us more practice. Also, can use Rouche's Theorem to find the roots. But to be honest, the lot of them would take me a while unless I really get the hang of it. Guess that's a bit of a contradiction since I added that bit about Riemann but I digress. Don't wait on me. Try and apply what we did above to those three. Mathematica is a program for doing math. Just like I typed in those commands and it output the roots. Tons more. Do a search on the web for Mathematica and also look down in the computer forum below. We have mathematica (and Matlab) problems that are solved. Everyone who does math should do Mathematica:

Code:

In[81]:= N[Solve[z^4 - 3 z^2 + 3 == 0, z], 4]
N[Solve[z^2 + I*z + 2 + I == 0, z], 4] 
N[Solve[z^7 + 6*z^3 + 7 == 0, z], 4]


Out[81]= {{z -> -1.2712 + 0.3406 I}, {z -> 
   1.2712 - 0.3406 I}, {z -> -1.2712 - 
    0.3406 I}, {z -> 1.2712 + 0.3406 I}}

Out[82]= {{z -> -0.3257 + 1.0350 I}, {z -> 
   0.326 - 2.035 I}}

Out[83]= {{z -> -1.000}, {z -> -1.122 - 
    1.201 I}, {z -> -1.122 + 1.201 I}, {z -> 
   0.4644 - 0.9559 I}, {z -> 
   0.4644 + 0.9559 I}, {z -> 
   1.157 - 0.978 I}, {z -> 1.157 + 0.978 I}}

**shawsend** · October 23rd 2008, 03:58 PM

Originally Posted by shadow_2145

So [69] is 2 zeros and [70] is 7 zeros? :/

Ok, I'm sorry. If you don't know Mathematica, than that output I'm showing is probably very cryptic. The input lines in my post above solved for the roots of the three equations. These roots are given in the form {{r1},{r2},...{rn}} for each Solve I executed. So the first output for the first equation yielded four roots:

{{z -> -1.2712 + 0.3406 I}, {z ->
1.2712 - 0.3406 I}, {z -> -1.2712 -
0.3406 I}, {z -> 1.2712 + 0.3406 I}}

First is -1.2712+0.3406i (given by that z-> notation)
Second is: 1.2712-0.3406i
Third is -1.2712-0.3406i
Fourth: 1.2712+0.3406i

You can interpret the rest.

**shadow_2145** · October 23rd 2008, 04:04 PM

Originally Posted by shawsend

Ok, I'm sorry. If you don't know Mathematica, than that output I'm showing is probably very cryptic. The input lines in my post above solved for the roots of the three equations. These roots are given in the form {{r1},{r2},...{rn}} for each Solve I executed. So the first output for the first equation yielded four roots:

{{z -> -1.2712 + 0.3406 I}, {z ->
1.2712 - 0.3406 I}, {z -> -1.2712 -
0.3406 I}, {z -> 1.2712 + 0.3406 I}}

First is -1.2712+0.3406i (given by that z-> notation)
Second is: 1.2712-0.3406i
Third is -1.2712-0.3406i
Fourth: 1.2712+0.3406i

You can interpret the rest.

First is -1.2712+0.3406i (given by that z-> notation)
Second is: 1.2712-0.3406i
Third is -1.2712-0.3406i
Fourth: 1.2712+0.3406i

So, it seems to me that the Fourth root is the only one in the first quadrant, so we have one zero. This cooresponds with the answer I got,

/hurray!

**shawsend** · October 23rd 2008, 04:21 PM

Ok good. Here's my work for #1: Call the legs of the contour as we go around them in the first quad, c1, c2, c3.

On c1 we have $R^4-3R^2+3$ and for large R, $\Delta_{c1}\arg f(z)=0$ (cake)

On c2 we have a similar situation as the example exercise:

$R^4e^{4it}\left(1-\frac{3}{R^2e^{2it}}+\frac{3}{R^4e^{2it}}\right)$
and using the same argument, we get $\Delta_{c2}\arg f(z)=2\pi$

However on c3, let z=yi so that $f(yi)=y^4+3y^2+3>0$ for all y so that $\Delta_{c3}\arg f=0$

So that over the entire contour, $\Delta_C\arg(f)=2\pi$ which means there is only a single root in this contour. Hey, I think I'm getting it too!

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