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 $-\cos(z)$.

  • $\begingroup$ The second is simply the definition of $cos(z)$ and $sin(z)$ on complex plane, right? $\endgroup$ – WaterBro Sep 29 '19 at 21:12
  • $\begingroup$ 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 $\sin(z)=\sum_{}\frac{(-1)^{n}x^{2n+1}}{(2n+1)!}$. $\endgroup$ – John Sep 30 '19 at 7:51

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