Induction of logarithmic derivatives of complex functions?

Question

I am trying to use induction to prove the logarithmic derivative of a complex function (called $P(Z)$ here).

I define a function $P(z) = (z-z_1)(z-z_2)...(z-z_n)$ and then I want to use induction on n to show that:

$P'(z)/P(z) = 1/(z-z_1) + 1/(z-z_2) + ... + 1/(z-z_n)$.

Note that the above notation is directly the logarithmic derivative of my defined function $P(z)$.

I'm having troubles finding a starting point for n, and then integrating my induction hypothesis into my additive step. Help?

Answer 1

I don't think you necessarily need to use induction here. Note that \begin{equation} \log P(z) = \log\left(\prod_{i=1}^{n}(z-z_i)\right) = \sum_{i=1}^{n}\log(z-z_i). \end{equation} It then follows that \begin{equation} P'(z)/P(z) = \frac{d}{dz}\log P(z) = \frac{d}{dz}\sum_{i=1}^{n}\log(z-z_i) = \sum_{i=1}^{n}\frac{d}{dz}\log(z-z_i) = \sum_{i=1}^{n}\frac{1}{z-z_i}. \end{equation} Note that since we have a finite sum we can safely move the derivative past the sum.

Answer 2

This is something that I did:

At $n=1$ We have $P(z) = (z-z_1)$ Then $P'(z) = 1$ Then $P'(z)/P(z) = 1/(z-z_1)$ which is true, so I assume it's true for k terms.

For K+1 terms: Let $Q(z) = P(z)*(z-z_{k+1})$ Then $Q'(z) = (P'z)(z-z_{k+1})+P(z)(1)$ Then doing some cancelling, we get $Q'(z)/Q(z) = P'(z)/P(z)+1/(z-z_{k+1})$, which completes the proof.

What do you think of that? (Wow, those symbols were a fingerful.)

Answer 3

The basis of the induction is trivial. So, let $$ P(z)=(z-z_1)(z-z_2)\dots(z-z_n)(z-z_{n+1}) $$ and $$ Q(z)=(z-z_1)(z-z_2)\dots(z-z_n) $$ so that $$ P(z)=Q(z)\cdot(z-z_{n+1}). $$ By the product rule $$ P'(z)=Q'(z)\cdot(z-z_{n+1})+Q(z) $$ and the induction hypothesis tells you that $$ Q'(z)=Q(z)\left(\frac{1}{z-z_1} + \frac{1}{z-z_2} + \dots + \frac{1}{z-z_n}\right). $$ Thus $$ \frac{P'(z)}{P(z)} = \frac{Q(z)}{P(z)}\left(\frac{1}{z-z_1} + \frac{1}{z-z_2} + \dots + \frac{1}{z-z_n}\right)(z-z_{n+1})+\frac{Q(z)}{P(z)} $$ Now apply $$ \frac{Q(z)}{P(z)}=\frac{1}{z-z_{n+1}} $$ and you're done.

Answer 4

$$P(z) = \prod_{r=1}^n z - z_r$$

$$\begin{align} \exp(P'(z)/P(z)) &= \sqrt[dz]{dP} \\ &= \sqrt[dz]{ \prod_{r=1}^n z - z_r}\\ &= \prod_{r=1}^n \sqrt[dz]{ z - z_r } \\ &= \prod_{r=1}^n\exp \frac{1}{ z - z_r} \\ \end{align}$$ $$\downarrow$$ $$\begin{align} P'(z)/P(z) &= \log\left(\prod_{r=1}^n \exp\frac{1}{ z - z_r}\right) \\ &= \sum_{r=1}^n \frac{1}{ z - z_r} \\ \end{align}$$

Pedantic: assuming $(\forall r)\,z_r$ is invariant of $z$