InterviewSolution
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InterviewSolution
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If `a+i b=(c+i)/(c-i)`, where `c`is real, prove that:`a^2+b^2=1a n d b/a=(2c)/(c^2-1)dot` |
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Answer» `a+ib = (c+i)/(c-i)` `=>a+ib = (c+i)/(c-i)**(c+i)/(c+i) = (c^2+i^2+2ci)/(c^2-i^2)` `=>a+ib = (c^2-1+2ci)/(c^2+1)` `=>a+ib = (c^2-1)/(c^2+1)+i(2c)/(c^2+1)` Comparing real and imaginary parts, `:. a = (c^2-1)/(c^2+1) and b = (2c)/(c^2+1)` (i) `L.H.S. = a^2+b^2 = [(c^2-1)/(c^2+1)]^2+ [(2c)/(c^2+1)]^2` `=(c^4+1-2c^2+4c^2)/(c^4+1+2c^2)` `=(c^4+1+2c^2)/(c^4+1+2c^2)` `=1 = R.H.S.` (ii) `L.H.S. = b/a = ((2c)/(c^2+1))/((c^2-1)/(c^2+1))` `=(2c)/(c^2-1) = R.H.S.` |
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