सिद्ध कीजिए - (i) `((1+i)/(1-i))^(200)=1` (ii) `((1-i)/(1+i))^(36)=1`

Simplify: ((1+i)/(1-i))^(200)

Simplify: ((1+i)/(1-i))^200

If ((1+i)/(1-i))^(3)-((1-i)/(1+i))^(3)=x+i y then (x, y)=

((1+i)/(1-i))^(4)+((1-i)/(1+i))^(4)=

1+(1+i)+(1+i)^(2)+(1+i)^(3)=

((1+i)/(1-i))^4+((1-i)/(1+i))^4=

Given that, z=(1+2i)/(1-i)=((1+2i)(1+i))/((1-i)(1+i))

Prove that: (i) (1-i)^(2)=-2i (ii) (1+i)^(4)xx(1+(1)/(i))^(4)=16 (iii) {i^(19)+((1)/(i))^(25)}^(2)=-4 (iv) i^(4n)+i^(4n+1)+i^(4n+2)+i^(4n+3)=0 (v) 2i^(2)+6i^(3)+3i^(16)-6i^(19)+4i^(25)=1+4i .

Find the modulus of (1+i)/(1-i) - (1-i)/(1+i)