Question

If a = cos x + i sin x, b = cos y + i sin b, c = cos z + i sin z and a+b+c = 0 then prove that 1/a + 1/b + 1/c = 0

Answer 1

Given,
a+b+c=0
⇒(cosx +isinx)+(cosy+isiny)+(cosz+isinz) = 0
⇒(cosx+cosy+cosz) +i(sinx+siny+sinz) = 0+i0
⇒(cosx+cosy+cosz) =0 or (sinx+siny+sinz) = 0

Now, 
1/a = 1/(cosx+isinx)
      =(cosx-isinx)/[(cosx+isinx)(cosx-isinx)]
      =(cosx-isinx)/(cos²x+sin²x)
      =(cosx-isinx)
similarly,
1/b = (cosy-isiny)
and , 1/z = (cosz-isinz)

∴(1/a +1/b +1/c)
=(cosx-isinx)+(cosy-isiny)+(cosz-isinz)
=(cosx+cosy+cosz) -i(sinx+siny+sinz)
=0