Misc 15 - Find modulus of (1 + i)/(1 - i) - (1 - i)/(1 + i)

Misc 15 Find the modulus of (1 + 𝑖)/(1 − 𝑖) − (1 − 𝑖)/(1 + 𝑖) First we solve (1 + 𝑖)/(1 − 𝑖) − (1 − 𝑖)/(1 + 𝑖) (1 + 𝑖)/(1 − 𝑖) − (1 − 𝑖)/(1 + 𝑖) = ((1 + 𝑖) × (1 + 𝑖) −(1 − 𝑖) × (1 − 𝑖))/((1−𝑖) (1+ 𝑖)) = ((1 + 𝑖)2 − ( 1 − 𝑖 )2)/((1 − 𝑖) (1 + 𝑖)) Using a2 – b2 = (a + b) (a – b) = ([( 1+ 𝑖 ) + ( 1 − 𝑖 )] [( 1+ 𝑖 ) − ( 1 − 𝑖 )])/((1)2 −(𝑖)2) = (( 1 + 1 + 𝑖 − 𝑖 ) ( 𝑖 + 𝑖 + 1 −1 ))/( 1 − 𝑖2) = (( 2 + 0 ) ( 2𝑖 + 0 ))/(1 − 𝑖2) = ((2) (2𝑖))/(1 −𝑖2) = 4𝑖/(1 −𝑖2) Putting i2 = - 1 = 4𝑖/( 1− ( − 1 ) ) = 4𝑖/( 1 + 1) = ( 4𝑖)/( 2) = 2𝑖 = 0 + 2𝑖 Thus , (1 + 𝑖)/(1 − 𝑖) − (1 − 𝑖)/(1 + 𝑖) = 0 + 2𝑖 Thus , (1 + 𝑖)/(1 − 𝑖) − (1 − 𝑖)/(1 + 𝑖) = 0 + 2𝑖 We need to find modulus of (1 + 𝑖)/(1 − 𝑖) − (1 − 𝑖)/(1 + 𝑖) Complex number (0 + 2𝑖) is of the form x + 𝑖y Hence x = 0 and y = 2 Modulus of z = |z| |z| = √(𝑥^2+𝑦2) = √((0)2+(2)2) = √(0+4) = √4 = 2 Hence, Modulus of (1 + 𝑖)/(1 − 𝑖) − (1 − 𝑖)/(1 + 𝑖) = 2