cosnθ+isinnθ is one of the value of (cosθ+isinθ)n when n=qp where q=1
Proof:
Case-1: Let n=0
Then the
L.H.S=(cosθ+isinθ)0=1
And the
R.H.S=cos0.θ+isin0.θ=1
Therefore L.H.S=R.H.S Case-2: Let n is a positive integer.
We prove it by the principal of mathematical induction .
For n=1
Then the
L.H.S=(cosθ+isinθ)1=cosθ+isinθ
And the
R.H.S=cosθ+isinθ
Therefore L.H.S=R.H.S
Let the statement is true for n=k, the we have (cosθ+isinθ)k=coskθ+isinkθ
Now
(cosθ+isinθ)k+1=(cosθ+isinθ)k.(cosθ+isinθ)=(coskθ+isinkθ).(cosθ+isinθ)=(coskθcosθ−sinkθsinθ)+i(coskθsinθ+sinkθcosθ)=cos(k+1)θ+isin(k+1)θ
Therefore the statement is true for n=k+1 when the statement is true for n=k. Then by the principal of mathematical induction the statement is true for n∈N.
Hence (cosθ+isinθ)n=cosnθ+isinnθ when n is a positive integer.Case-3: Let n is a negative integer.
Let m=−n, then m is positive ineteger.
Then by Case-2, we have
(cosθ+isinθ)m=cosmθ+isinmθ
Therefore
(cosθ+isinθ)n=(cosθ+isinθ)−m=(cosθ+isinθ)m1=(cosmθ+isinmθ)1=cosmθ−isinmθ=cos(−m)θ+isin(−m)θ=cosnθ+isinnθ
Hence (cosθ+isinθ)n=cosnθ+isinnθ when n is a negative integer.Case-4: Let n=qp where and q=1 .(cosqpθ+isinqpθ)q=(cosq.qpθ+isinq.qpθ)byCase−2=(cospθ+isinpθ)=(cosθ+isinθ)pbyCase−2or3
Therefore
(cosqpθ+isinqpθ)q=(cosθ+isinθ)p
Implies (cosqpθ+isinqpθ) is one of the value of (cosθ+isinθ)qp
That is (cosθ+isinθ)n=cosnθ+isinnθ when n=qp and q=1.