Question

Find: The real part: \[\Re[(4+i)e ^{-(1+i)x}]\]

Answer 1

I know the real part of 4+i is just 4, but I have no idea about the exp{...} part. Is there a way to rewrite it?

Answer 2

So find the real part of e^(-i) and e^(-x) separately and multiply them together?

Answer 3

the real part of e^(-x) is e^(-x)
(I think they would use z for the variable name if they meant x to be complex)

so if we expand
(4 +i) exp(-x) exp(-i)= 4 exp(-x) ( sin(-x )+ i cos(-x)) + i *( sin(-x )+ i cos(-x))

Answer 4

Also, isn't it exp(-x-xi)?

Answer 5

yes, I dropped an x . ...

Answer 6

now we can take the real part...

simplify sin(-x)= -sin(x)
and cos(-x) = cos(x)

Answer 7

I'm looking over this, but I'm lost on the sin(-x )+ i cos(-x)) + i *( sin(-x )+ i cos(-x) part? How did you get that since e^(-i) expands to cos(-1)+isin(-1)

Answer 8

Got it! But now it seems even more complicated. How would you find the real part from all of that?

Answer 9

wait... I should use the correct identity:

exp(ix)= cos(x) + i sin(x)

Answer 10

Let me re-type it correctly

(4+i) * exp(-(1+i)x) = (4+i)* exp(-x) * exp(-ix)

= 4 exp(-x) * exp(-ix) + i exp(-x) * exp(-ix)

now rewrite
exp(-ix)= cos(-x) + i sin(-x) = cos(x) - i sin(x)

Answer 11

I get a real part of
4 exp(-x) cos(x) + exp(-x) sin(x)

Answer 12

So now it's \[4e ^{-x}(\cos x-i \ \sin x)+ie ^{-x}(\cos x-i \ \sin x)\] How did you find the real part from that?

Answer 13

the first term has a real part of 4 exp(-x) cos(x) (and an imaginary part -i * 4 exp(-x) sin(x))

Answer 14

in other words, look for A+B i where A and B are pure reals, and select the A part

Answer 15

the second term has a real part -i*i*exp(-x) sin(x) = exp(-x) sin(x)

Answer 16

Got it! I checked the second part and it's perfect! Thank you so much! I couldn't seem to find much about this online, so it really helped!

Answer 17

yw... though it was a bit of a struggle for me to get the details correct...

Answer 18

I have one other problem like this I'm stuck on now. It's:
\[\Im[ \frac{3i}{2-i}e ^{(2+i)x}]\]
Rewriting the exponential part: exp(2x)exp(ix) => exp(2x)cos(x)+exp(2x)isin(x)

And multiplying by the fraction I got:
\[ \frac{3ie ^{2x}\cos(x) -3e ^{2x}\sin(x)}{2-i}\] But I'm not sure what to do from here

Answer 19

I would clear the denominator by multiplying top and bottom by the complex conjugate 2+i

Answer 20

Alright I got:
\[\frac{-6e ^{2x}\sin(x)-3e ^{2x}\cos(x)+i[6e ^{2x}\cos(x)-3e ^{2x}\sin(x)]}{5}\] So it would just be the imaginary part of that?

Answer 21

yes, looks good

Answer 22

Thanks!