Multiplicative inverse of the complex number $$(\sin \theta, \cos \theta)$$ is

$$\sum_\limits{k=0}^{440} i^k=x+i y \Rightarrow x^{100}+x^{99} y+x^{242} y^2+x^{97} y^3=$$

If $$e^{i \theta}=\operatorname{cis} \theta$$, then $$\sum_\limits{n=0}^{\infty} \frac{\cos (n \theta)}{2^n}=$$

$$i z^3+z^2-z+i=0 \Rightarrow|z|=$$