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#1 Posted 09:41am 17-01-10 complex nos.the number of ordered pair (a,b) such that (a + ib)^2010 = a - ib is??? |
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#2 Posted 10:24am 17-01-10 Re: complex nos.(o,1)? |
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#3 Posted 10:28am 17-01-10 Re: complex nos.i asked the NUMBER of ordered PAIRS.... |
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#4 Posted 3:26pm 17-01-10 Re: complex nos.Taking modulus on both sides we see that [im]http://codecogs.izyba.com/gif.latex?%28a%5E2+b%5E2%29%5E%7B2010%7D%20%3D%20a%5E2+b%5E2[/im] so that [im]http://codecogs.izyba.com/gif.latex?a%5E2+b%5E2%20%3D%200[/im] or [im]http://codecogs.izyba.com/gif.latex?a%5E2+b%5E2%20%3D%201[/im] So a=b=0 is a solution. Otherwise let z=a+ib where |z| =1. Then a-ib = 1/z Hence our equation is [im]http://codecogs.izyba.com/gif.latex?z%5E%7B2011%7D%3D1[/im] which has 2011 distinct solutions. Hence total number of ordered pairs is 2012 |
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