Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.

? asked in Science & MathematicsMathematics · 1 decade ago

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

  • 1 decade ago
    Favorite 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).

  • lavena
    Lv 4
    5 years ago

    "returned interior the day whilst there grew to become into patriotism nevertheless interior the rustic people like those activists could have been kept away from via the respectable people and labled treasonous." what? whilst mccarthyism ruled washington? "we've lost this patriotism and something is going despite the fact that if it is going against the rustic, in actuality those varieties of concept's have become very huge-unfold. the difficulty is those idiots that try this type of outrage do no longer care how damaging it incredibly is and ***** then as quickly as we've grow to be a undesirable u . s .." ok, whilst did excepting and swallowing blatant lies grow to be patriotism and questioning those lies grow to be damaging? do you no longer think of that bush mendacity is efficient? "My question to you tho, it is effective those activists say those issues approximately 9/eleven and deliver those video's to the troops and that they might because of the fact they have freedom of speech, why then does not Reagan have a similar freedoms?" the adaptation being, we attempt to give up a re-enactment of the crusades, and that pretend republican (a better classification being corporatist) is inquiring for, advacating and screaming for homicide! there's a music that announces, and that i quote, "a number of individuals who choose forces are a similar that burn crosses." you look to sound such as you belong to that group, rachel...

Still have questions? Get your answers by asking now.