# Factorization of $(1+2i)z^2+6iz+2i-1$ to find the residues, but it doesn't work

I'm doing complex analysis and I have the following $$\int_{C}\frac{2dz}{(2i+1)z^2+6iz+2i-1}$$ where $C$ is the unit circle. I tried factorizing this polynomial and find the residues but I really can't solve this.

This is my work: $$z_{1,2} = \frac{-3i\pm \sqrt{-9-(-4-1)} }{2i+1}=\frac{-3i\pm 2i}{1+2i} \cdot \frac{1-2i}{1-2i}= \frac{-3i-6\pm 2i\pm 4}{5}$$ which gives $z_1 = -\frac{2}{5}-\frac{i}{5}$ and $z_2 = -2-i$. Now clearly $z_2$ is outside the unit circle so I just have to consider the residue of $z_1$.

However I think I'm either making a factorisation mistake or a mistake in the roots. Indeed if we check the roots, I don't get the same polynomial! $$(z+2+i)\left(z+\frac{2}{5}+\frac{i}{5}\right) = z^2 +z\left(\frac{12}{5}+\frac{6i}{5}\right)+\frac{4i}{5}+\frac{3}{5}$$ which is clearly not the same polynomial I started with. Also if I try to calculate the residue using the limit, I don't get the correct one. Where's my mistake?

EDIT

As requested, I'll add some more calculations. To calculate the residue I did the following: $$\lim_{z\to z_1}(z+\frac{2}{5}+\frac{i}{5})\frac{2}{(z+2+i)\left(z+\frac{2}{5}+\frac{i}{5}\right)} = \frac{2}{\frac{8}{5}+\frac{4i}{5}} = \frac{5}{4+2i} = \frac{20-10i}{20}=1-\frac{i}{2}$$ which is clearly wrong as this integral is the mapping of a real integral of the first type (rational function in sine and cosine). Indeed by Cauchy's Residue Theorem this should give that the value of the initial real integral is $2\pi i (1-\frac{i}{2}) = 2\pi i+\pi$ which is a complex value, so this is wrong

• Multiply through by the coefficient of the $z^2$ term to check the polynomial. – Michael Burr Dec 29 '16 at 14:25
• @MichaelBurr do you mean $(2i+1)$? I know this is a stupid question, however how come I don't get the correct polynomial by finding the roots? – Euler_Salter Dec 29 '16 at 14:26
• @Euler_Salter Of course, if $p(x)$ is a polynomial, then $\lambda p(x)$ is another polynomial with the same roots for any nonzero scalar $\lambda$... but does multiplying with $2i+1$ even fix the mistake above? – rschwieb Dec 29 '16 at 14:27
• @MichaelBurr and is there a quicker way to factorize these kind of polynomials? – Euler_Salter Dec 29 '16 at 14:28
• @rschwieb true. So whenever I have a polynomial with a coefficient of $z^2$ different from $1$ or $0$, I have to find the roots and after that I multiply the polynomial $(z-z_1)(z-z_2)$ by that coefficient? – Euler_Salter Dec 29 '16 at 14:29

Let's start with the factorization question. The given integrand is a fraction, and to use the residue theorem, we must find the roots of the denominator, i.e., the roots of $$(2i+1)z^2+6iz+(2i-1).$$ To find the roots of this polynomial, it's easiest to use the quadratic formula to get that the roots are \begin{align*} \frac{-6i\pm\sqrt{(6i)^2-4(2i+1)(2i-1)}}{2(2i+1)}&=\frac{-6i\pm\sqrt{-36+20}}{4i+2}\\ &=\frac{-6i\pm\sqrt{(6i)^2-4(2i+1)(2i-1)}}{2(2i+1)}\\ &=\frac{-6i\pm\sqrt{-16}}{4i+2}\\ &=\frac{-6i\pm4i}{4i+2} \end{align*} Therefore, the roots are $\frac{-2i}{4i+2}=\frac{-i}{2i+1}$ and $\frac{-5i}{2i+1}$. Multiplying through by the conjugate of the denominator, $(1-2i)$ gives that the roots are $\left(-\frac{2}{5}-\frac{1}{5}i\right)$ and $(-2-i)$.

As the OP observes, the polynomial $$\left(z-\left(-\frac{2}{5}-\frac{1}{5}i\right)\right)(z-(-2-i))=z^2+\left(\frac{12}{5}+\frac{6}{5}i\right)z+\left(\frac{3}{5}+\frac{4}{5}i\right)$$ is not the original denominator. This is another polynomial with the same roots, as the original polynomial. To make the polynomials the same, we should multiply by $(1+2i)$ to correct the coefficient of $z^2$. In fact $$(1+2i)\left(z^2+\left(\frac{12}{5}+\frac{6}{5}i\right)z+\left(\frac{3}{5}+\frac{4}{5}i\right)\right)=(2i+1)z^2+6iz+(2i-1),$$ the original polynomial.

Therefore, the original polynomial factors as $$(2i+1)z^2+6iz+(2i-1)=(2i+1)\left(z-\left(-\frac{2}{5}-\frac{1}{5}i\right)\right)(z-(-2-i)).$$

To deal with the residues, the OP correctly determines that $-2-i$ is outside the unit circle, so it doesn't matter for the integral (although it could be used to calculate the integral, but that is a story for another time).

Therefore, the value of the given integral is $2\pi i$ times the residue at $\left(-\frac{2}{5}-\frac{1}{5}i\right)$. Rewriting the integral as $$\int_\gamma \frac{2}{(1+2i)(z-(-2-i))}\cdot\frac{1}{z-\left(-\frac{2}{5}-\frac{1}{5}i\right)}dz,$$ the value of the residue can be calculated by substituting $\left(-\frac{2}{5}-\frac{1}{5}i\right)$ for $z$ in $\frac{2}{(1+2i)(z-(-2-i))}$. This substitution results in $-\frac{1}{2}i$ and Cauchy's residue theorem gives that the integral is $\pi$.

The only mistake that the OP seems to make throughout is forgetting the factor of $(1+2i)$ when factoring the denominator. When that is included, all of the answers (and the limit approach in the question) agree.

• perfect I'd say! – Euler_Salter Dec 29 '16 at 15:10

If you were only interested in the value of your integral $f$ you could let $z=\exp (i \theta )$ and get

$$fr=\int_0^{2 \pi } \frac{1}{\sin (\theta )+2 \cos (\theta )+3} \, d\theta$$

In order to calculate the integral we attempt to use the fundamental theorem of calculus, i.e. taking the difference of the anitiderivative at both ends of the integration interval. But this holds only for continuous antiderivatives. If the antiderivative contains jumps these must be taken into account.

An antiderivative of the integrand is

$$fa=-\tan ^{-1}\left(\frac{2}{\tan \left(\frac{\theta }{2}\right)+1}\right)$$

In the region of integration the denominator vanishes at $\theta =\frac{3 \pi }{2}$. Here the antiderivative makes a jump from

$$\lim_{\theta \to \left(\frac{3 \pi }{2}\right)^-} \, fa=\frac{\pi }{2}$$

to

$$\lim_{\theta \to \left(\frac{3 \pi }{2}\right)^+} \, fa=-\frac{\pi }{2}$$

i.e. it jumps by an amount of $\pi$.

Now the antiderivative at both ends of the integration interval is equal to $-\tan ^{-1}(2)$. Hence the difference is zero, and we are left with the jump which gives $f=fr=\pi$.

EDIT

I'm afraid I have made it more complicated than necessary.

Let us therefore apply a "standard" procedure and do this for the more general integral (displayed here in original form and reduced form)

$$\int_0^{2 \pi } \frac{1}{a+b \cos (\theta )+c \sin (\theta )} \, d\theta =\int_0^{2 \pi } \frac{1}{a+\sqrt{b^2+c^2} \cos (\phi )} \, d\phi =\frac{2 \pi }{\sqrt{a^2-b^2-c^2}}$$

which holds for $a^2-b^2-c^2>0$

Making the "standard" substitution $\theta \to 2 \tan ^{-1}(t)$, $\text{d$\theta $}=\frac{2 \text{dt}}{t^2+1}$ and splitting the integral into the two parts from $0$ to $\pi$ and form $\pi$ to $2 \pi$ gives the complete integration range in to from $-\infty$ to $\infty$.

We have

$$\cos (\theta )=\cos ^2\left(\frac{\theta }{2}\right)-\sin ^2\left(\frac{\theta }{2}\right)=\left(1-t^2\right) \cos ^2\left(\frac{\theta }{2}\right)$$ $$\sin (\theta )=2 \sin \left(\frac{\theta }{2}\right) \cos \left(\frac{\theta }{2}\right)=2 t \cos ^2\left(\frac{\theta }{2}\right)$$ $$\frac{1}{\cos ^2\left(\frac{\theta }{2}\right)}=t^2+1$$

The integral becomes

$$2 \int_{-\infty }^{\infty } \frac{1}{a \left(t^2+1\right)+b \left(1-t^2\right)+2 c t} \, dt$$

Supplementing the square, shifting the integration varibale (this makes no dfference in our infnte interval) and extracting factors gives finally

$$\frac{\int_{-\infty }^{\infty } \frac{2}{v^2+1} \, dv}{\sqrt{a^2-b^2-c^2}}=\frac{2 \pi }{\sqrt{a^2-b^2-c^2}}$$

• you lost me half way but this looks like an elegant way of solving it – Euler_Salter Dec 29 '16 at 16:29
• See my 2nd EDIT – Dr. Wolfgang Hintze Dec 29 '16 at 20:09
• One sort of shortcut: periodicity and the angle sum formula (amongst other things) tell you that $\int_0^{2\pi} f(a+b\cos\theta+c\sin\theta)\,d\theta = \int_0^{2\pi} f(a+\sqrt{b^2+c^2}\cos\phi)\,d\phi$, and playing around with the homogeneity in $a,b,c$ means that the only integral you really need to compute is $\int_0^{2\pi}\frac{1}{1+\varepsilon \cos \psi}\,d\psi$ for $|\varepsilon|<1$. – πr8 Dec 29 '16 at 20:14
• @πr8 You are right. This was originally in my text. I have dropped it for brevity, but have now reinserted it. – Dr. Wolfgang Hintze Dec 29 '16 at 21:20