Why does $(e^{-\frac{3}{4}t}-e^{it})(\frac{3}{4}-i) = e^{-\frac{3}{4}t}-\cos t+ \frac{3}{4}\sin t$

37 Views Asked by At

I read this while going through some derivations in my mathematical physics class. I thought of multiplying everything and applying Euler's formula but I don't understand how the $ie^{-\frac{3}{4}t}$ could be turned into something real.

A little more context: this was after doing convolution with the Laplace transform and then doing something with real numbers to solve this deceptively easy-looking problem $y''+y'+\frac{3}{16}y(t) = \sin (t),\quad y(0) = y'(0) = 0$.

1

There are 1 best solutions below

0
On

Your simplification is not correct. use $e^{it}=\cos t+ i\sin t$, then $$(e^{-3t/4}-\cos t-i \sin t)(\frac{3}{4}-i)=\frac{3}{4} e^{-3t/4}-\frac{3}{4} \cos t- \frac{3i}{4}\sin t-i 3^{-3t/4}-i \cos t -\sin t.$$