Summation of imaginary numbers

Question

At !!theta=6!! and !!Z=e^(itheta)!!,

!!sum_(m=1)\^15 Im(Z^(2m-1))=?!!

!!a)" "1/(2Sin6)!!

!!b)" "1/(Sin6)!!

!!c)" "1/(2Sin12)!!

!!d)" "1/(Sin12)!!

Answer

We can solve this complex equation using more than one method.

Method 1

We can see that if we will put value of !!m!! in the summation, we will get GP series.

We can easily calculate summation of GP and then separate imaginary terms.

!!sum_(m=1)\^15 Im(Z^(2m-1))!!

!!= Im(Z+Z^3+Z^5+.......Z^29)!!

!!= Im{ Z(1-(Z^2)^15)/(1-Z^2) }!!

!!= Im{ (1-Z^30)/(1/Z-Z) }!!

Putting the value of !!Z=e^(itheta)= (Costheta + iSintheta)!!

!!= Im{ (1- Cos30theta - iSin30theta )/(-2iSintheta) }!!

Multiplying numerator and denominator with !!i!!, we get

!!= Im{ i(1- Cos30theta - iSin30theta )/(2Sintheta) }!!

!!= Im{ i(1- Cos30theta) + Sin30theta )/(2Sintheta) }!!

Now taking out imaginary term, we get

!!=(1- Cos30theta)/(2Sintheta)!!

Putting the value of !!theta = 6!!, we get

!!= (1- Cos30xx6)/(2Sin6)!!

!!= (1- Cos180)/(2Sin6)!!

!!= (1- (-1))/(2Sin6)!!

!!= 2/(2Sin6)!!

!!= 1/Sin6!!

Method 2

!!sum_(m=1)\^15 Im(Z^(2m-1))!!

In the second method we can initially seperate imaginary term after expanding the summation.

We get,

!!Sintheta + Sin3theta + Sin5theta +........+ Sin29theta!!

This summation can be solved using trinometric formula:

!!Sina + Sin(a+b) + Sin(a+2b)..... + Sin(a+(n-1b)) = ((Sin(nb)/2)/(Sinb/2))Sin(a+(n-1)b/2)!!

hence,

!!= ((Sin15*2theta/2)/(Sintheta/2))xxSin(theta+(15-1)2theta/2)!!

!!= ((Sin15theta)/(Sintheta))xxSin15theta!!

!!= (Sin15theta)^2/Sintheta!!

Putting the value of !!theta = 6!!, we get

!!= (Sin15xx6)^2/Sin6!!

!!= (Sin90)^2/Sin6!!

!!= 1/Sin6!!

Get it on Google Play