If a+ib = c+i / c-i, where c is real, prove that a^{2} + b^{2 }=1 and b/a = 2c / c^{2} -1. Share with your friends Share 0 Shanul answered this Here is the link for your queryhttps://www.meritnation.com/ask-answer/question/if-a-ib-c-i-c-i-where-c-is-real-part-prove-that-a-square-.../complex-numbers-and-quadratic-equations/2830157 1 View Full Answer Shri Agiwal answered this it is given that a +ib = c+i / c-i . mod or taking magnitudes on both sides : root a+ib^{2} = [root c+i / rootc-1] ^{2} ==> root a^{2} + b^{2} = 1 squaring on both sides , a^{2} + b^{2} = 1 to bring the complex number in std form , we can multiply both nr and dr by c+i ==> (c+i)^{2} / c^{2} + 1 ==> c^{2}+2ci -1 / c^{2} +1 ==> a = c^{2} - 1/ c^{2}+1 , b = 2c / c^{2} +1 ==> b / a = 2c / c^{2}-1 ( c^{2}+1 get cancelled )hope u got it 5 Lakshita answered this a+ib = c+i / c-i= c+i * c+i / c-i * c+i=c^{2}+i^{2}+2ci / c^{2}-i^{2}=c^{2}-1+2ci / c^{2}+1a=c^{2}-1 / c^{2}+1 b=2c / c^{2}+1b/a= 2c /c^{2}+1 * c^{2}+1 / c^{2}-1 b/a= 2c / c^{2}-1 (as c^{2}+1 cut off)a^{2}+b^{2} = (c^{2}-1/c^{2}+1)^{2 }+ (2c/c^{2}+1)^{2}=c^{4}+1-2c^{2} / (c^{2}+1) + 4c^{2} /( c^{2}+1)^{2}=c^{4}+1-2c^{2}+4c^{2} / (c^{2}+1)^{2}=c^{4}+1+2c^{2} / (c^{2}+1)^{2}=(c^{2}+1)^{2 }/ (c^{2}+1)^{2 }= 1 =a^{2}+b^{2} 34