Find the real and imaginary part of sin(4+3i)

  • Thread starter Thread starter tatiana_eggs
  • Start date Start date
  • Tags Tags
    Imaginary
AI Thread Summary
To find the real and imaginary parts of sin(4+3i), the user initially applied the sine addition formula and converted terms into exponential form. They encountered difficulties in simplifying the resulting expressions and questioned whether a calculator was necessary for further calculations. A helpful suggestion was made to utilize Euler's formula, leading to a breakthrough in solving the problem. Ultimately, the user successfully derived the real and imaginary components of sin(4+3i). This approach highlights the importance of using exponential identities in complex analysis.
tatiana_eggs
Messages
69
Reaction score
0

Homework Statement



Find the real and imaginary part of sin(4+3i)

Homework Equations



sinx = \frac{e^z - e^(-z)}{2i}

cosx = \frac{e^z + e^(-z)}{2}

sin(iy) = i\frac{e^y - e^(-y)}{2}

cos(iy) = \frac{e^y + e^(-y)}{2}

various trig identities

The Attempt at a Solution



So I used sin(x+y) trig identity and got
sin4*cos3i + sin3i*cos4

I turned them all into exponents using the appropriate equations stated in (2).

I got to a point where nothing is really calculable by hand/head. Is there an easier way to do this or does the calculator need to be used at a certain point to calculate the real part(terms grouped w/o i) and the imaginary part (terms grouped with i).

If so, then I guess I need help getting the terms grouped together to calculate the real and imaginary parts.

Where I am stuck is at:

\frac{e^{3+4i}+e^{-3+4i}-e^{3-4i}+e^{-3-4i}}{4i} +
\frac{e^{3+4i}-e^{-3+4i}+e^{3-4i}-e^{-3-4i}}{4}

(the two fractions should be added together)

Now what should I do with all these lovely exponents? Should I have even gone this route?
 
Physics news on Phys.org
tatiana_eggs said:

Homework Statement



Find the real and imaginary part of sin(4+3i)

Homework Equations



sinx = \frac{e^z - e^(-z)}{2i}

cosx = \frac{e^z + e^(-z)}{2}

sin(iy) = i\frac{e^y - e^(-y)}{2}

cos(iy) = \frac{e^y + e^(-y)}{2}

various trig identities

The Attempt at a Solution



So I used sin(x+y) trig identity and got
sin4*cos3i + sin3i*cos4

I turned them all into exponents using the appropriate equations stated in (2).

I got to a point where nothing is really calculable by hand/head. Is there an easier way to do this or does the calculator need to be used at a certain point to calculate the real part(terms grouped w/o i) and the imaginary part (terms grouped with i).

If so, then I guess I need help getting the terms grouped together to calculate the real and imaginary parts.

Where I am stuck is at:

\frac{e^{3+4i}+e^{-3+4i}-e^{3-4i}+e^{-3-4i}}{4i} +
\frac{e^{3+4i}-e^{-3+4i}+e^{3-4i}-e^{-3-4i}}{4}

(the two fractions should be added together)

Now what should I do with all these lovely exponents? Should I have even gone this route?
Now use e^{3+ 4i}= e^3cos(4)+ i e^3sin(4), etc.
 
That was just the hint I needed, Halls. Thanks! Finally got it.
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...
Back
Top