Question

Find the Modulus of 1+I/1-i - 1-i /1+i

Answer 1

1+i/1-i - 1-i/1+i
= Taking LCM of both sides
=(1+i)^2 - (1-i)^2/(1-i)(1+i)
= 1+i^2+2i-1-i^2+2i / 1^2 -(i^2)
= Putting value of i^2
= 1-1+2i-1+1+2i / 1+1
= 4i / 2
=0+2i
Modulus of complex no is underrot x^2+y^2
: where x is real part and y is img part.
Z= underoot (0^2) + (2)^2
= underoot 0+4
= underoot 4
=2
This is the answr .
May this help uh

Answer 2

$\huge\star{\red{\underline{\mathfrak{Answer}}}}$

1+i/1-i - 1-i/1+i

{Taking LCM of both sides}

=(1+i)^2 - (1-i)^2/(1-i)(1+i)

= 1+i^2+2i-1-i^2+2i / 1^2 -(i^2)

{Putting value of i^2}

= 1-1+2i-1+1+2i / 1+1

= 4i / 2

=0+2i

Modulus of complex no is underrot x^2+y^2: where x is real part and y is img part.

Z= underoot (0^2) + (2)^2

= underoot 0+4

= underoot 4

=2