# how to prove that ${P^{'}}$ has no real root?

Let $P(z)$ be a monic polynomial with complex coefficients with all roots distinct and in $\{z \in C : \Im(z) \lt 0\}$.

$(a)$ Prove that the sum of all the residues of $\frac{P^{'}}{P}$ is the degree of the polynomial $P$.

$(b)$ Prove that $P^{'}$ has no real root.

My idea was that option $(a)$ as I take $f(z)$=$\frac{p^{'}(z)}{p(z)}$

$\deg(p(z))\ge \deg(p^{'}(z))+2$

Residue theorem: If $f$ is analytic in a domain except for isolated singularities at $a_1,\dots a_k$ then for any closed contour $\gamma\in D$ on which none of the points $a_k$ lie, we have $$\frac{1}{2\pi i}\int_{\gamma}f(z)dz=\sum_{1}^{k}n(\gamma;a_k)Res[f(z);a_k].$$

Here I don't know how to proceed further.

For option $(b)$ if I take even polynomial degree that $p(x) =x^2+1$ then it will not have real roots

As I don't know the actual proof. Please assist and help me.

• pliz help me,,anybody????
– user396850
Commented Feb 28, 2018 at 13:55
• Part a) is an application of the so-called argument principle (see en.wikipedia.org/wiki/Argument_principle), since $$\sum \operatorname{Res} (\frac{p'}{p}) = \frac{1}{2 \pi i} \int_\gamma \frac{p'}{p} \ dz = \text{Number of zeros of } p$$ which equals $\operatorname{deg}(p)$ by the fundamental theorem of algebra. For the contour $\gamma$, you can take an sufficiently large half circle in the lower half plane around the origin. Commented Feb 28, 2018 at 13:56
• thanks a lots complexFlo and For part B) ????
– user396850
Commented Feb 28, 2018 at 13:59
• There is a general result that says that the roots of $P'$ are contained in any half plane that contains the roots of $P$. Here, you are asked to prove a special case. Are you aware of that general result? Commented Feb 28, 2018 at 16:03
• Find the Gauss-Lucas theorem on wikipedia. Commented Feb 28, 2018 at 16:22

The first part is a residue theorem exercise. For the second part, write $p(z) = \prod (z-a_i)$, where the $a_i$ are the roots of $p$, possibly with repetition. Then $p'(z)/p(z) = \sum \frac{1}{z-a_i}$. Suppose $z$ is any complex number with $\Im(z) \ge 0$. Then $$\Im(p'(z)/p(z)) = \Im \left(\sum \frac{1}{z-a_i}\right) = \Im\left(\sum \frac{\bar z - \bar a_i}{|z-a_i|^2}\right).$$ Now since $\Im(z) \ge 0$, it follows that $\Im(z - a_i) > 0\$ for all $i$, hence $\Im(\bar z - \bar a_i) < 0$ for all $i$, hence $\Im(p'(z)/p(z)) \ne 0$, hence $p'(z) \ne 0$.