If a+ib = c+i / c-i, where c is real, prove that a2 + b2 =1 and b/a = 2c / c2 -1.

  • 1

 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 

  • 5

 a+ib = c+i / c-i

= c+i * c+i / c-i * c+i

=c2+i2+2ci / c2-i2

=c2-1+2ci / c2+1

a=c2-1 / c2+1     b=2c / c2+1

b/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)

=  1 =a2+b2

  • 34
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