I'm refreshing my memory on sines, cosines, SOHCAHTOA, and writing complex numbers in polar form – I chose the complex number $3+4i$ to try and write it in polar form. The modulus is easily computed to be $5$. However, I'm having some trouble with computing its argument. Based on drawing out the right triangle, thinking of $\tan(\theta)$, then the argument must be $\tan^{-1} (\frac{4}{3})$.
However, how do we actually compute this?
I don't think it's $\frac{1}{\tan(\theta)}$ = $\frac{\cos(\theta)}{\sin(\theta)}$, because that just seems weird and more like cotangent.
Where have I gone wrong?
Thanks,
Best Answer
Just a complement to other answers. At first, we have that $\arctan\frac{4}{3}$ is not a rational multiple of $\pi$. Assuming $\arctan\frac{4}{3}\in\pi\mathbb{Q}$ we have that $z=\frac{3+4i}{5}$ is a primitive $q$-th root of unity, but its minimal polynomial over $\mathbb{Q}$ is given by $5z^2-6z+5$, that is not a monic polynomial, hence is not a cyclotomic polynomial. So $\arctan\frac{4}{3}$ is not a "nice" angle. However, there are plenty of ways for approximating values of the arctangent function, for instance by exploiting continued fractions, the (D'Aurizio)-Shafer-Fink inequality, Machin-like formulas or Taylor series. For instance, since
$$ \arctan(z)=\frac{\pi }{4}+\frac{1}{2}(z-1)-\frac{1}{4} (z-1)^2+\frac{1}{12} (z-1)^3-\frac{1}{40} (z-1)^5+\frac{1}{48} (z-1)^6+O((z-1)^7)$$ we have: $$ \arctan\frac{4}{3}\approx \frac{\pi}{4}+\frac{24827}{174960} $$ and this approximation gives us the first five figures $\color{red}{0.92729}$.