Find the sum of a series involving imaginary numbers

[leilsilver]

Find the sum of cos 45 + icos135 +...+i^n cos (45+90n)+...+ i^40 cos 3645.

I tried putting a sequence or a series, but it didn't quite work, because I had trouble with the i along with the cosign. I know that it keeps on rotating at the 45 degrees points, but just with the i added into it. I can do it the longhand way...(which is drawing a circle and adding numbers...) But can anyone find a shortcut to it?

 

[galactus]

Maybe try the addition formula and work from there.

\(\displaystyle \L\\cos(45+90n)=cos(45)cos(90n)-sin(45)sin(90n)\)

\(\displaystyle \L\\\frac{1}{\sqrt{2}}\sum_{n=0}^{40}(i^{n}cos(90n))-\frac{1}{\sqrt{2}}\sum_{n=0}^{40}(i^{n}sin(90n))\)

 

[leilsilver]

Thanks, I will....

 

[pka]

Building on galactus’ hint, take note:
\(\displaystyle \L
\cos \left( {\left( {2k + 1} \right)\pi } \right) = 0\quad \& \quad \cos \left( {\left( {2k} \right)\pi } \right) = \pm 1\).
Likewise \(\displaystyle \L
\sin \left( {\left( {2k + 1} \right)\pi /2} \right) = \pm 1\quad \& \quad \sin \left( {\left( {2k} \right)\pi } \right) = 0\).

Therefore: \(\displaystyle \L
\sum\limits_{k = 0}^{40} {i^k \cos (\pi /4 + \pi k/2)} = \left( {1/\sqrt 2 } \right)\sum\limits_{k = 0}^{20} {\left( { - 1} \right)^k } \cos \left( {k\pi } \right) - i\left( {1/\sqrt 2 } \right)\sum\limits_{k = 0}^{19} {\left( { - 1} \right)^k } \sin \left( {\left( {2k + 1} \right)\pi /2} \right)\)

BTW. I do not use degrees, so changed to proper notation.

 

[leilsilver]

I don't get what you're trying to explain to me pka. I understood galactus' hint but you kind of lost me. Is your equation the substitute for the 90n?

 

[pka]

I was trying to help you understand why the answer is:
\(\displaystyle \L
\frac{{21}}{{\sqrt 2 }} - i\frac{{20}}{{\sqrt 2 }}\).

 

[leilsilver]

Got it