Question

find the modulus and amplitude of 1+i÷1-i

Answer 1

modulus of 1+i÷1-i
=/√1^2+1^2 ÷ √1^2+(-1)^2/
=√2 ÷ √2=/1/
= 1
amplitude of 1+i÷1-i
=tan^-1 /(y/x)/
=tan^-1 (/(1/1)/ / /(1/-1)/)
=tan^-1 (/(1)/ / /(-1)/)
=π/2

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