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Asked by devendra.mittal | 7th Nov, 2009, 08:33: PM

Expert Answer:

Let z = r (cosθ + isinθ)

Then, for any positive integer n,  zn = rn (cosnθ + isinnθ)

Now, LHS = |Im(zn)| = |rn sinnθ| = rn sinnθ

RHS = n|Im(z)||z|n-1 =  n |r sinθ|(r2(cos2θ + sin2θ))n-1

                                    = nrsinθ rn-1(1)n-1

                                   = nrnsinθ

According to the question we need to prove sinnθ ≤ nsinθ

We will prove this by using mathematical induction.

P(n):  sinnθ ≤ nsinθ

P(1): sinθ = (1)sinθ

Let P(k) be true for any integer k, Then

sinkθ ≤ ksinθ  .............(1)

Now, we need to prove that the relation is true for any (k+1).

i.e sin(k+1)θ ≤ (k+1)sinθ

⇒ sinkθ cosθ + coskθsinθ ≤ (k+1)sinθ

⇒ ksinθcosθ + coskθsinθ ≤ (k+1)sinθ

⇒sinθ (kcosθ + + coskθ) ≤ (k+1)sinθ .......(2)

Now, we know that

-1≤ cosθ ≤1

-k ≤ kcosθ ≤k

-k + coskθ ≤ kcosθ + + coskθ  ≤ k + coskθ

coskθ ≤1

⇒kcosθ + + coskθ  ≤k +1 ...........(3)

Now using the result of equation (3) in (2) we get

sin(k+1)θ ≤ (k+1)sinθ

Hence, the answer.

Answered by  | 12th Jan, 2010, 04:31: PM

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