The complex form of the equation for an ellipse with foci at 1 and -1 is $|z-1|+|z+1|=\sqrt{8}$.
a) Find the values of $a$ and $b$ such that $x^2/a^2+y^2/b^2=1$ describe the same ellipse.
b) Let $C$ be the quarter of the ellipse starting at the right most point and ending at the top most point (one quarter of the ellipse). Calculate $\int_C \cos z\ dz$.
For part a) I was able to obtain the values $a=\sqrt{2}$ and $b=1$. For part b) I paramterized the ellipse as $z(t)=\sqrt{2}\cos t+i\sin(t)$, and therefore $\dot{z}(t)=-\sqrt{2}\sin t+i\cos t$, and the integral (I think) should be
$\int_C \cos z\ dz=\int_0^{\pi/2}f(z(t))\dot{z}(t)\ dt=\int_0^{\pi/2}\cos(\sqrt{2}\cos t+i\sin t)\cdot(-\sqrt{2}\sin t+i\cot t)\ dt.$
But I can't evaluate that integral for the life of me. Another trick I tried was the "Fundamental Theorem" and I got
$\int_C\cos z\ dz=\sin(i)-\sin(\sqrt{2})$,
which I also cannot evaluate. Can somebody help me find my mistake or lead me toward the solution? Thanks.
As the comment says, the Newton-Leibniz theorem is the way.
The power series $1+z+\frac{z^2}{2!}+\frac{z^3}{3!}+\dots$ converge not only for all real $z$, but also for all complex $z$. This defines the exponential function $e^z$. It turns out that if $z$ is purely imaginary, then $e^z$ is moving on the unit circle of the complex plane, moreover $$e^{i\varphi}=\cos\varphi+i\,\sin\varphi\,.$$ From it one can get the power series of $\cos$ and $\sin$ simultaneously, or we can express from above that $$\cos\varphi=\frac{e^{i\varphi}+e^{-i\varphi}}2,\quad\quad \sin\varphi=\frac{e^{i\varphi}-e^{-i\varphi}}{2i}\,.$$ But, the same way as above, these expressions also make sense when $\varphi\in\Bbb C$.
Namely, we get $\cos(i\,x)=\cosh(x)$ and $\sin(i\,x)=i\sinh(x)$.