If [(1-i)/(1+i)]100 =a+ib, then find (a,b)

[(1-i)/(1+i)]100 =a+ib

Lets rationalise it.

[(1 - i)(1 - i) / (1 + i)(1 - i)]100 =a+ib

[ (1 - i)2 / (1  - i2  ) ]100  = a + ib

[(1 + i2 - 2i) / 2 ]100  = a + ib

[-2i/2]100  = a + ib

(-i)100  = a + ib

[(-i)2 ]50    = a + ib

(-1)50  = a + ib

1  = a + ib

a  = 1

b  = 0

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