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# How to show the complex equation P(z) = exp(wz) has finitely many roots on every line?

Let w be complex constant and P a non constant polynomial with complex coefficients. Show that, on every line of the complex plane, the equation P(z) = exp(wz) has finitely many roots.

Does this remain true for w ≠ 0 if we allow z to run over the entire complex plane?

Thank you

### 2 Answers

- SteinerLv 71 decade agoFavorite Answer
For the first question, it suffices to prove the claim for the real axis. To see this, observe that any line L on C can be described by the equation z = z_0 + t z_1, where z_0 is a pure imaginary (the intersection of L and the imaginary axis), z1 is a unit complex in the direction of L and t is a real variable from - to + infinity. Exactly like the vector equation of a line on R².

With some elementary complex algebra, you can show there's a bijection between the roots of P(z) = exp(wz) on L and the roots of a similar equation for real z. I leave the details to you.

So, supposing z is real and w = a + bi, algebraic form, we have exp(wz) = exp(az) exp(bzi) and |exp(wz)| = exp(az). In addition, the magnitude of |P(z)| is the same as the magnitude of |z|^n, where n ≥ 1 is the degree of P. We have 3 cases:

1. a > 0

We readily see that |exp(az)| - |P(az)| goes to ∞ as z → ∞ and to -∞ as z → -∞. Hence, for sufficenly large positive z and for negative z with sufficiently large |z|, we must have ezp(wz) ≠ P(z). This shows that the roots of exp(wz) = P(z) are in a bounded interval of the real axis and, therefore, form a bounded (possibly empty) set. If this set is infinite, then it has a limit point on R and, since exponential and polynomials are entire functions, the properties of complex functions imply we must have exp(wz) = P(z) everywhere on C. But the simple behavior of these functions on the real axis rules this out. We can also see this observing that the derivatives of any polynomial eventually become identically 0, which is never the case of an exponential.

So, for a > 0, the claim is true

a = 0

Then, |exp(wz)| = 1 for every real z. Since |P(z)| → ∞ as z goes to +or - infinity, a simliar reasoning to the one of (a) shows we have finitely many roots.

a < 0

In this case, |exp(az) - P(az)| goes to -∞ as z & rarr; ∞ and to ∞ as z → -∞. It's the dual of case (a). Exactly the same reasonig shows the equation has finitely many roots on the real line.

So, we are done for real z and, therefore, for any line on C.

Now, suppose w ≠ 0 and let z run over the entire complex plane. We have infinitely many roots. To see this, put f(z) = exp(-wz) P(z). Then, f is entire and its roots are exactly the roots of P. Since P is a polynomial, f has finitely many roots. In addition, f is not a polynomial. It it were, then exp would be either a rational or a polynomial function. In the 1st case, it wouldn't be entire, contrarily to what is a fact; in the second case, we'd get a the same contadiction we mentioned before. Now, recall Picard's Theorem:

If f is entire and non polynomial, then, with possible exception of a single complex z_0, all other complexes are attained by f infinitely many times.

Our f is entire and non polynomial. And since it attains 0 only finitely many times, 0 is, for our f, the exception of Picard's theorem. This means that every non zero complex is attained by f infinitely many times, so that we have f(z) = exp(-wz) P(z) = 1 for infinitely many complexes z. This is the same as to say that exp(wz) = P(z) has infinitely many roots (countably many, because we have entire functions)

We see that, for w ≠ 0, on the entire plane we have infinitely many roots even if P is constant (and non identically 0, if you consider the zero function is polynomial).

- lavenaLv 45 years ago
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