# Solving a higher derivative using the complex eponential

I am currently trying to solve:

$$\left(\frac{d^{61}}{dt^{61}}\right)(e^{-t+1} cos⁡(t))$$

So far I've converted cos(t) into Re($$e^{it}$$) and thus got Re($$e^{(-1+i)t+1}$$). This is where I've gotten a bit stuck. I tried separating it out into Re($$e^{(-1+i)t}e^{1}$$) and then using the product rule to solve that, but I'm not sure whether that's valid, and if it is, where to go from there in solving the problem.

• Welcome to Math.SE! Did you intend to write $=$ there? Sep 6, 2020 at 5:35
• @KevinLópezAquino Nope, sorry! I've gotten rid of it now. Sep 6, 2020 at 5:51
• You can do that, or just write 2cosx = e^ix + e^-ix Sep 6, 2020 at 5:55
• The $61$-st derivative of $e^{(-1+i)t}$ is $(-1+i)^{61}e^{(-1+i)t}$. Sep 6, 2020 at 6:15

## 1 Answer

We can continue on the track that you have. Using the fact that

$$\frac{d^n}{dt^n}e^{\lambda t} = \lambda^n e^{\lambda t}$$

we get

$$e\frac{d^{61}}{dt^{61}}e^{(-1+i)t} = e\cdot(-1+i)^{61}e^{(-1+i)t}$$

Convert the complex number to polar form to compute the power

$$-1+i = \sqrt{2}e^{i\frac{3\pi}{4}} \implies (-1+i)^{61} = 2^{\frac{61}{2}}e^{i\frac{183\pi}{4}} = -2^{30}\cdot\sqrt{2} e^{i\frac{3\pi}{4}}=2^{30}(1-i)$$

which means we can rewrite the complex derivative as

$$e^{-t+1}\cdot 2^{30}\cdot(1-i)\cdot(\cos t + i\sin t)$$ Taking the real part gives us the final answer

$$\frac{d^{61}}{dt^{61}}e^{-t+1}\cos t = 2^{30}e^{-t+1}(\cos t+\sin t)$$