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CBSE Class 11-science Answered

(cos?+sin?)^n=cos n?+isin n?
Asked by narendra | 02 Mar, 2014, 10:28: AM
answered-by-expert Expert Answer
DeMoivre'sTheorem: If z is any complex number, z equals r open parentheses cos theta plus i sin theta close parentheses and n is a natural number, then z to the power of n equals open square brackets r open parentheses cos theta plus i sin theta close parentheses close square brackets to the power of n equals r to the power of n open square brackets cos n theta plus i sin n theta close square brackets
Let us prove this above theorem by the method of induction.
 
W h e n space n equals 1 comma space w e space h a v e comma space z to the power of 1 equals open square brackets r to the power of 1 open parentheses cos theta plus sin theta close parentheses to the power of 1 close square brackets equals r open parentheses cos theta plus sin theta close parentheses T h u s comma space t h e space t h e o r e m space i s space t r u e space f o r space n equals 1.
 
Now let us assume that the theorem is true for n = k and we need to prove that the theorem is true for n = k + 1.
 
Since the theorem is true for n = k, we have,
 
z to the power of k equals open square brackets r open parentheses cos theta plus i sin theta close parentheses close square brackets to the power of k equals r to the power of k open square brackets cos k theta plus i sin k theta close square brackets
Consider
z to the power of k plus 1 end exponent equals z to the power of k cross times z to the power of 1 equals z to the power of k cross times z equals r to the power of k open parentheses cos k theta plus i sin k theta close parentheses cross times r open parentheses cos theta plus i sin theta close parentheses equals r to the power of k cross times r cross times open square brackets cos open parentheses k theta plus theta close parentheses plus i sin open parentheses k theta plus theta close parentheses close square brackets equals z to the power of k plus 1 end exponent cross times open parentheses cos open parentheses k plus 1 close parentheses theta plus i sin open parentheses k plus 1 close parentheses theta close parentheses
 
Thus the theorem is true for n = k + 1.
 
Therefore, the theorem is proved.
Answered by | 02 Mar, 2014, 05:08: PM

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