# Thread: complex numbers and derivatives

1. ## complex numbers and derivatives

Hi all,

I need some help with the following question:

the 257th derivative of e^(-t)*sint

When i do it, i get an answer of 2 ^(257/2)*i*(cost(t) + i*sin(t)) but the answer is 2^128*e^(-t)*(cos(t) - sin(t))

ArTiCk

2. Originally Posted by ArTiCK
Hi all,

I need some help with the following question:

the 257th derivative of e^(-t)*sint

When i do it, i get an answer of 2 ^(257/2)*i*(cost(t) + i*sin(t)) but the answer is 2^128*e^(-t)*(cos(t) - sin(t))

ArTiCk
You know that $\displaystyle e^{-t} \sin t = \text{Im} [e^{-t} e^{it}] = \text{Im}[e^{(-1+i)t}]$.

I leave it to you to therefore justify that $\displaystyle \frac{d^{257}}{dt^{257}}\left( e^{-t} \sin t\right) = \text{Im} \left[ \frac{d^{257}}{dt^{257}}\left( e^{(-1+i)t} \right) \right]$.

So the answer is $\displaystyle \text{Im} \left[ (-1 + i)^{257} e^{(-1+i)t} \right] = e^{-t} \, \text{Im} \left[ (-1 + i)^{257} e^{it} \right]$.

Note that

$\displaystyle -1 + i = \sqrt{2} \, e^{3 \pi/4}$

$\displaystyle \Rightarrow (-1 + i)^{257} = 2^{257/2} \, e^{771\pi/4} = 2^{257/2} \, e^{3\pi/4}$

$\displaystyle = 2^{257/2} \left( \cos \frac{3 \pi}{4} + i \sin \frac{3 \pi}{4} \right) = 2^{257/2} \left(-\frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}} \right) = 2^{256/2} (-1 + i) = 2^{128} (-1 + i)$.

Therefore the answer will be $\displaystyle 2^{128} e^{-t} \, \text{Im} \left[ (-1 + i) e^{it} \right] = 2^{128} e^{-t} \, \text{Im} \left[ (-1 + i) (\cos t + i \sin t) \right] = ........$