If a+ib = c+i / c-i, where c is real, prove that a2 + b2 =1 and b/a = 2c / c2 -1. Share with your friends Share 1 shanul.singh 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 2 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+ib2 = [root c+i / rootc-1] 2 ==> root a2 + b2 = 1 squaring on both sides , a2 + b2 = 1 to bring the complex number in std form , we can multiply both nr and dr by c+i ==> (c+i)2 / c2 + 1 ==> c2+2ci -1 / c2 +1 ==> a = c2 - 1/ c2+1 , b = 2c / c2 +1 ==> b / a = 2c / c2-1 ( c2+1 get cancelled )hope u got it 4 Lakshita answered this a+ib = c+i / c-i= c+i * c+i / c-i * c+i=c2+i2+2ci / c2-i2=c2-1+2ci / c2+1a=c2-1 / c2+1 b=2c / c2+1b/a= 2c /c2+1 * c2+1 / c2-1 b/a= 2c / c2-1 (as c2+1 cut off)a2+b2 = (c2-1/c2+1)2 + (2c/c2+1)2=c4+1-2c2 / (c2+1) + 4c2 /( c2+1)2=c4+1-2c2+4c2 / (c2+1)2=c4+1+2c2 / (c2+1)2=(c2+1)2 / (c2+1)2 = 1 =a2+b2 37
