CBSE Class 11-science Answered

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 r^n-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.