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Question Number 32160 by rahul 19 last updated on 20/Mar/18

If z=cosθ+isinθ is a root of equation a_0 z^n +a_1 z^(n−1) +a_2 z^(n−2) +.....+a_(n−1) z+a_n =0 then prove that: i) a_0 +a_1 cos θ+a_2 cos 2θ+.....+a_n cos nθ=0 ii) a_1 sin θ + a_2 sin 2θ+....+a_n sin nθ=0.

$$\boldsymbol{{I}}{f}\:{z}={cos}\theta+{isin}\theta\:{is}\:{a}\:{root}\:{of}\:{equation} \\ $$$${a}_{\mathrm{0}} {z}^{{n}} +{a}_{\mathrm{1}} {z}^{{n}−\mathrm{1}} +{a}_{\mathrm{2}} {z}^{{n}−\mathrm{2}} +.....+{a}_{{n}−\mathrm{1}} {z}+{a}_{{n}} =\mathrm{0} \\ $$$${then}\:{prove}\:{that}: \\ $$$$\left.{i}\right)\:{a}_{\mathrm{0}} +{a}_{\mathrm{1}} \mathrm{cos}\:\theta+{a}_{\mathrm{2}} \mathrm{cos}\:\mathrm{2}\theta+.....+{a}_{{n}} \mathrm{cos}\:{n}\theta=\mathrm{0} \\ $$$$\left.{ii}\right)\:{a}_{\mathrm{1}} \mathrm{sin}\:\theta\:+\:{a}_{\mathrm{2}} \mathrm{sin}\:\mathrm{2}\theta+....+{a}_{{n}} \mathrm{sin}\:{n}\theta=\mathrm{0}. \\ $$

Answered by rahul 19 last updated on 20/Mar/18

My try: Dividing the whole eq^n by z^n , ⇒a_0 +a_1 z^(−1) +.......+a_(n−1) z^(1−n) +a_n z^(−n) =0 Now z satisfies the above eq^n , ⇒a_0 +a_1 (cos θ−isin θ)+........... ⇒...+a_(n−1) (cos(n−1)θ−isin(n−1)θ) ⇒+a_n (cosnθ−isin nθ)=0. (by de moivre′s theorem). Collecting the real and imaginary part seperately , ⇒(a_0 +a_1 cos θ+a_2 cos 2θ+.....+a_n cos nθ) −i(a_1 sin θ+a_2 sin 2θ+.....+a_n sin nθ)=0 How to proceed further ?

$${My}\:{try}: \\ $$$${Dividing}\:{the}\:{whole}\:{eq}^{{n}} \:{by}\:{z}^{{n}} , \\ $$$$\Rightarrow{a}_{\mathrm{0}} +{a}_{\mathrm{1}} {z}^{−\mathrm{1}} +.......+{a}_{{n}−\mathrm{1}} {z}^{\mathrm{1}−{n}} +{a}_{{n}} {z}^{−{n}} =\mathrm{0} \\ $$$${Now}\:{z}\:{satisfies}\:{the}\:{above}\:{eq}^{{n}} , \\ $$$$\Rightarrow{a}_{\mathrm{0}} +{a}_{\mathrm{1}} \left(\mathrm{cos}\:\theta−{i}\mathrm{sin}\:\theta\right)+........... \\ $$$$\Rightarrow...+{a}_{{n}−\mathrm{1}} \left({cos}\left({n}−\mathrm{1}\right)\theta−{isin}\left({n}−\mathrm{1}\right)\theta\right) \\ $$$$\Rightarrow+{a}_{{n}} \left({cosn}\theta−{isin}\:{n}\theta\right)=\mathrm{0}. \\ $$$$\left({by}\:{de}\:{moivre}'{s}\:{theorem}\right). \\ $$$$\boldsymbol{{C}}{ollecting}\:{the}\:{real}\:{and}\:{imaginary} \\ $$$${part}\:{seperately}\:, \\ $$$$\Rightarrow\left({a}_{\mathrm{0}} +{a}_{\mathrm{1}} \mathrm{cos}\:\theta+{a}_{\mathrm{2}} \mathrm{cos}\:\mathrm{2}\theta+.....+{a}_{{n}} \mathrm{cos}\:{n}\theta\right) \\ $$$$−{i}\left({a}_{\mathrm{1}} \mathrm{sin}\:\theta+{a}_{\mathrm{2}} \mathrm{sin}\:\mathrm{2}\theta+.....+{a}_{{n}} \mathrm{sin}\:{n}\theta\right)=\mathrm{0} \\ $$$${How}\:{to}\:{proceed}\:{further}\:? \\ $$

Commented by rahul 19 last updated on 21/Mar/18

hey, @above but here my no. is a−ib=0⇒a=ib so is it compulsory that a=b=0.

$${hey},\:@{above}\:{but}\:{here}\:{my}\:{no}.\:{is} \\ $$$${a}−{ib}=\mathrm{0}\Rightarrow{a}={ib}\:{so}\:{is}\:{it}\:{compulsory}\:{that} \\ $$$${a}={b}=\mathrm{0}. \\ $$

Commented by Tinkutara last updated on 20/Mar/18

A complex number a+ib=0 only when a=b=0. So you have done complete solution!

$${A}\:{complex}\:{number}\:{a}+{ib}=\mathrm{0}\:{only}\:{when} \\ $$$${a}={b}=\mathrm{0}.\:{So}\:{you}\:{have}\:{done}\:{complete} \\ $$$${solution}! \\ $$

Commented by rahul 19 last updated on 20/Mar/18

Hahaha..... thanks for checking!

$${Hahaha}..... \\ $$$${thanks}\:{for}\:{checking}! \\ $$

Commented by mrW2 last updated on 21/Mar/18

in the notation for complex numbers z=a+ib the sign + doesn′t mean the sum of a and ib.

$${in}\:{the}\:{notation}\:{for}\:{complex}\:{numbers} \\ $$$${z}={a}+{ib} \\ $$$${the}\:{sign}\:+\:{doesn}'{t}\:{mean}\:{the}\:{sum}\:{of} \\ $$$${a}\:{and}\:{ib}. \\ $$

Commented by mrW2 last updated on 21/Mar/18

Commented by rahul 19 last updated on 21/Mar/18

ok, thank you sir.

$${ok},\:{thank}\:{you}\:{sir}. \\ $$

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