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)

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