If z=ei θ and (3 cos 3 θ+2 cos 2 θ+5 cos 5 θ/3 sin 3 θ+2 sin 2 θ+5 sin 5 θ)=(i displaystyle∑r=010 ar zr/ displaystyle∑r=010 br zr) then (( displaystyle∑r=010 ar+ displaystyle∑r=010 br)/10)=

Question

If z=ei θ and (3 cos 3 θ+2 cos 2 θ+5 cos 5 θ/3 sin 3 θ+2 sin 2 θ+5 sin 5 θ)=(i displaystyle∑r=010 ar zr/ displaystyle∑r=010 br zr) then (( displaystyle∑r=010 ar+ displaystyle∑r=010 br)/10)= (A) 0 (B) 1 (C) 2 (D) 3

Tardigrade Admin · Accepted Answer

For a complex number Z=ei θ it is given that (3 cos 3 θ+2 cos 2 θ+5 cos 5 θ/3 sin 3 θ+2 sin 2 θ+5 sin 5 θ)=i ( displaystyle∑r=010 ar Zr/ displaystyle∑r=010 br Zr) ⇒ ( displaystyle∑r=010 ar zr+ displaystyle∑r=010 br zr/ displaystyle∑r=010 ar zr- displaystyle∑r=010 br zr)=(3 ei 3 θ+2 ei 2 θ+5 ei 5 θ/3 e-i 3 θ+2 e-i 2 θ+5 e-i 5 θ) ⇒ ( displaystyle∑r=010 zr(ar+br)/ displaystyle∑r=010 zr(ar-br))=(3 ei 3 θ+2 ei 2 θ+5 ei 5 θ/3 e-i 3 θ+2 e-i 2 θ+5 e-i 5 θ) ⇒ a0+b0=a1+b1=a4+b4=a6+b6 =a7+b7=a8+b8 =a9+b9=a10+b10=0 and a2+b2=2, a3+b3=3, a5+b5=5 ∴ ( displaystyle∑r=010(ar+br)/10)=(2+3+5/10)=1

Q. If z=eiθ and 3sin3θ+2sin2θ+5sin5θ3cos3θ+2cos2θ+5cos5θ​=r=0∑10​br​zrir=0∑10​ar​zr​ then 10(r=0∑10​ar​+r=0∑10​br​)​=

0

1

2

3

Q. If
$z = e^{i θ}$ and $\frac{3 c o s 3 θ + 2 c o s 2 θ + 5 c o s 5 θ}{3 s i n 3 θ + 2 s i n 2 θ + 5 s i n 5 θ} = \frac{i r = 0 \sum 10 a _{r} z ^{r}}{r = 0 \sum 10 b _{r} z ^{r}}$ then
$\frac{( r = 0 \sum 10 a _{r} + r = 0 \sum 10 b _{r} )}{10} =$