# Why is $\frac{\cos^2(z)-1}{z^2}$ at $z=0$ a removable singularity and not a simple pole?

I've been learning singularities in a complex analysis class, and I was given a problem to determine the kind of singularity for the function above. My first thought was to factorize the above numerator to get $$\cos^2(z) - 1 = (\cos(z) - 1)(\cos(z) + 1)$$. Therefore the numerator would have a single root, and the denominator would have a double root, meaning overall that $$z=0$$ would be a simple pole of the equation. According to my professor, this is wrong and the singularity is actually removable, and I should find the limiting value.

After finding this out, I made three solutions to find the limiting value. These being:

$$\lim_{z\to0}\frac{\cos^2(z)-1}{z^2}$$ = $$\lim_{z\to0}\frac{-\sin^2(z)}{z^2}$$ = $$\lim_{z\to0}\frac{-\sin(z)}{z}$$ * $$\lim_{z\to0}\frac{\sin(z)}{z}$$ = $$-1$$.

$$\lim_{z\to0}\frac{\cos^2(z)-1}{z^2}$$ = $$\lim_{z\to0}\frac{(1-\frac{z^2}{2!}+\frac{z^4}{4!}-...)^2-1}{z^2}$$ = $$\lim_{z\to0}\frac{(1-{z^2}+...)-1}{z^2}$$ = $$\lim_{z\to0}\frac{-z^2+...}{z^2}$$ = $$-1$$ via Taylor Expansions.

$$\lim_{z\to0}\frac{\cos^2(z)-1}{z^2}$$ = $$\lim_{z\to0}\frac{-2\sin(z)\cos(z)}{2z}$$ = $$\lim_{z\to0}\frac{2\sin^2(z)-2\cos^2(z)}{2}$$ = $$-1$$ via L'Hopital's Rule.

What I don't understand is what made my original solution incorrect in the first place, does my factorization not work in $$\mathbb{C}$$? Or is what I attempted to do not a valid way of determining singularities. Any help would be appreciated, thank you!

• “Therefore the numerator would have a single root ...” – No, $(\cos(z)-1)$ has a double zero at the origin. Commented Jan 19 at 11:58
• Thank you, I don't know how I managed to miss something like that! I forgot that only works with polynomials. Commented Jan 19 at 12:03

$$\cos(z) -1$$ has a zero of order $$2$$ at $$0$$. $$\cos(z) -1=-\frac{z^{2}}{2!}+\frac{z^{4}}{4!}-....$$