Question

Question #a53f5

Answer 1

`Z=1+0i`

Explanation:

We have `(1+i)/(1-i)+2/(1+i)`.

We must get a common denominator. This is `(1-i)(1+i)`.

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

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

`=i+1-i`

`=1`, or `1+0i`

Answer 2

See the answer below...

Explanation:

`Z=(1+i)/(1-i)+2/(1+i)`

`=>Z=((1+i)^2+2(1-i))/(1-i^2)" "["simple addition"]`

`=>Z=(1+2i+i^2+2-2i)/(1-i^2)`

`=>Z=(1+2i-1+2-2i)/(1+1)" "[i^2=-1]`

`=>Z=1`

`=>color(red)(ul(bar(|color(green)(Z=1+0 cdot i)|`

Hope it helps...
Thank you..

Answer 3

`1`

Explanation:

You need to find a common denominator, add and simplify (understanding that `i^2=-1` )

`Z=(1+i)/(1-i)+2/(1+i)`

`Z=(1+i)/(1-i)xx(1+i)/(1+i)+2/(1+i)xx(1-i)/(1-i)`

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

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

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

`Z=(3+i^2)/(1+1)`

`Z=2/2`

`Z=1`