# Find primitives ( = anti-derivatives) of the functions $sin(z)$ and $cos(z)$.

Find primitives ( = anti-derivatives) of the functions $$sin(z)$$ and $$cos(z)$$.

Since both functions are holomorphic, we can certainly use their power series expression to get their primitives, also, in $$\Bbb R$$, we know that $$sin(x)$$ has anti-derivative of $$-cos(x)$$ and $$cos(x)$$ has $$sin(x)$$. But in complex plane, where even $$\frac{1}{z}$$ does not have a primitive, can I directly say that they have primitives of $$-cos(z)$$ and $$sin(z)$$? If cannot, how can I find an explicit formula for the primitive without using power series?

In complex plan, $$\cos(z)$$ and $$\sin(z)$$ are defined by there power series, so sure, your argument as valid. An other possible argument is $$\cos(z)=\frac{e^{iz}+e^{-iz}}{2}\quad \text{and}\quad \sin(z)=\frac{e^{iz}-e^{-iz}}{2i}.$$ If you know that the primitive of $$e^{iz}$$ is $$\frac{1}{i}e^{iz}$$, then you can also easily conclude that the primitive of $$\cos(z)$$ is $$\sin(z)$$ and the primitive of $$\sin(z)$$ is $$-\cos(z)$$.
• The second is simply the definition of $cos(z)$ and $sin(z)$ on complex plane, right? – WaterBro Sep 29 '19 at 21:12
• For me, the fact that $e^{iz}=\cos(z)+i\sin(z)$ really come from the fact that $\cos(z)=\sum_{}\frac{(-1)^{n}x^{2n}}{(2n)!}$ and $\sin(z)=\sum_{}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}$. – John Sep 30 '19 at 7:51