No Complex Analysis Equation z^4 + z + 5 = 0

In summary, the conversation discusses different methods for showing that the equation z^4 + z + 5 = 0 has no solution in the set {z is a subset of C: modulus of z is less than 1}. The methods mentioned include using Rouché's Theorem, De Moivre's theorem, and the triangle inequality. The conversation also briefly touches on the topic of interchangeable limits in integration and the concept of homotopy in computing path integrals.
  • #1
Hi all,,
I have a problems on complex Analysis:
Show that the equation
z^4 + z + 5 = 0 has no solution in the set { z is a subset of C: modulus of z is less than 1}

i tried doing it using Triangle inequality although i got it but i am looking for a better solution...Pls help
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  • #2
You could use Rouché's Theorem:

If f,g are analytic functions on a simply connected region G and C is a contour in G such that |g(z)|<|f(z)| for all points z on C, then f+g has as many roots inside C as f.

Using this, we note that for |z|=1 we have:|z^4+z|<=2<5.
So the z^4+z+5 has as many roots inside the circle {z:|z|<1 } as 5. So that's clearly zero.
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  • #3
I don't know Rouche's theorem, my knowledge of complex analysis is not great. But here's how I would do it.

Let [tex]z = r(\cos\theta + i\sin\theta)[/tex]

where r is the real magnitude and theta the argument.

Assume a solution for z with r < 1 exists.

Then using De Moivre's theorem and splitting off the real part of the equation gives (I'm skipping a few steps because I'm a bit rushed, you can fill in the gaps) :

[tex]r^4\cos{(4\theta)} + r\cos\theta = -5[/tex]

Taking the magnitude of the LHS,

[tex]|LHS| \leq |r^4\cos{(4\theta)}| + |r\cos\theta| < |r\cos{(4\theta)}| + |r\cos\theta|[/tex]

since for nonnegative r < 1, [itex]r^4 < r[/itex]

[tex]|r\cos{(4\theta)}| + |r\cos\theta| = r(|\cos{(4\theta)}| + |\cos\theta|) < |\cos{(4\theta)}| + |\cos\theta| < 2[/tex]

using the boundedness of the trig ratios.

So |LHS| < 2, but for the orig. eqn to be satisfied, |LHS| = 5, this is a contradiction (RAA, QED).
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  • #4
Curious solution is quite simple
  • #5
Thx Curious ..
Galileo can u illustrate more about urs solution...May be i can learn a new method to do such preoblems
  • #6
I admit Rouché's theorem is overkill for this problem, but since you asked for different methods I posted it anyway.
As Curious also showed, it can easily be done with the triangle inequality, simply because for |z|<1 we have |z^4+z|<2.

I don't know any elementary way to prove Rouché's theorem. Only one using the principle of the argument involving contour integration. But I can show some nice illustrations:

- Take [itex]P(z)=z^n+a_{n-1}z^{n-1}+...+a_1z+a_0[/itex]. Intuitively we would expect |z^n| to grow faster than [itex]|a_{n-1}z^{n-1}+...+a_1z+a_0|[/itex] (like the real counterpart). So let [itex]f(z)=z^n[/itex] and [itex]g(z)=a_{n-1}z^{n-1}+...+a_1z+a_0[/itex], so that P(z)=f(z)+g(z). If we take the circle |z|=R, then:


[tex]|g(z)| \leq |a_{n-1}|R^{n-1}+|a_{n-2}|R^{n-2}+...+|a_1|R+|a_0|[/tex]


[tex]\frac{|g(z)|}{|f(z)|}\leq \frac{|a_{n-1}|}{R}+\frac{|a_{n-2}|}{R^2}+\frac{|a_{n-3}|}{R^3}+...+\frac{|a_1|}{R^{n-1}}+\frac{|a_0|}{R^n}[/tex]

The right lid goes to zero if [itex]R\to \infty[/itex], so for R large enough we have |g(z)|<|f(z)| on the circle |z|=R.
Now Rouché's theorem says that P(z) has as many roots inside this circle as f(z)=z^n, which is n. This gives a nonstandard proof of the fundamental theorem of algebra.

-You can approximate the annular region where roots are located. Take [itex]P(z)=z^4+3z+1[/itex]. Then for the circle |z|=1 we have [itex]|z^4+1|\leq 2<3=|3z|[/itex], so P(z) has as many roots inside |z|=1 as 3z. That is, it has 1 root inside this circle.
For |z|=2 we have [itex]|3z+1|\leq 7 <16 =|z^4|[/itex], so there are 4 roots of P(z) (all of them) inside the circle |z|=2.
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  • #7
Thx very much Galileo...i understood it is very well explained.
  • #8
here is another question in understanding the proof of which i am completely Lost:
Pls help me

Let f(z)=Sigma(n=0 to infinity)a(base-n)z^n be a power series with radius of convergence R>0.Show that
f is differentiable on U with deivative f'(z)=Sigma(n=1 to infinity)n a(base-n)z^n-1 and f' has radius of convergence of R.

I will be highly thankful to you..
  • #9
There is a famous result stating when you can interchange summation and differentiation. Try googling for it.
  • #10
by summation u mean here integration...actually i did not get the proper heading for googling...
It all presented vague results
Does interchange of limits take place here
  • #12
thx for link Matt,,
That i know that term by term differentiation gives the derivative but My teacher said that its natural to accept this fact but not obvious...There is a trivial Proof::
The proof she did i could not grasp
  • #13
path integration is obviously a basic tool. Hence basic properties of path integration are also fundamental.

The first question is: over which paths does an integrand have the same value? this is basic for computing integrals by simplifying the path.

The fundamental result is that if the differential form you are integrating is "closed", i.e. satisfies the mixed partials condition, [i.e. the "curl" is zero], then the the integral is the same over any two "homotopic" paths, i.e. paths that are deformable continuously into each other.

all holomorphic functions and hence all polynomials f(z), define closed one forms f(z)dz.

The hypothesis of rouche's "theorem" [more of a remark] guarantees existence of an obvious homotopy along straight lines between the two paths.

More generally, the integral of a closed form is the same along any two 'homologous" paths, which is the commutative version of homotopy. e.g. any 1 form, closed or not, has integral zero over the path A*B *A^-1 *B^-1, even when this path is not contractible.

Geometrically, and intuitively, two paths are homologous when their "difference" forms the boundary of some surface

verbum sapienti: Learning this "homotopy" technique for computing path integrals will do you much more good in the long run, and short run, than solving the trivial problem at hand.

remark: the correspondence between integrals of closed forms and boundary properties of loops, gives rise to "de rham" cohomology. I.e. a plane region has "holes" in it, if and only if there exist closed differential one forms with non zero integrals around some closed path in the region.

thus the analytic invariant: {closed diff one forms}/{forms of type df}, measure the geometry - topology of the region.
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  • #14
mathwonk that's cool

Related to No Complex Analysis Equation z^4 + z + 5 = 0

1. What is a complex analysis equation?

A complex analysis equation is an equation that involves complex numbers, which are numbers that contain both a real and imaginary component. These equations are used to study the properties and behavior of functions with complex inputs and outputs.

2. What is the solution to the equation z^4 + z + 5 = 0?

The solution to this equation is a complex number that satisfies the equation when plugged in for z. This equation has four solutions, known as roots, in the complex plane.

3. How can I solve this equation without using complex analysis?

This equation can be solved using algebraic techniques, such as factoring or the quadratic formula. However, these methods will only give the real solutions to the equation and will not take into account the complex roots.

4. What is the significance of the z^4 term in this equation?

The z^4 term in this equation represents the highest power of z in the equation. This indicates that the equation is a quartic equation, which can have up to four complex solutions.

5. Can this equation be graphed on a complex plane?

Yes, this equation can be graphed on a complex plane, with the real component of z on the horizontal axis and the imaginary component on the vertical axis. The solutions to the equation will be represented by points on the graph where the function crosses the x-axis.

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