Question #a53f5

3 Answers
Feb 10, 2018

Z=1+0iZ=1+0i

Explanation:

We have (1+i)/(1-i)+2/(1+i)1+i1i+21+i.

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

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

=(2i)/2+(2(1-i))/2=2i2+2(1i)2

=i+1-i=i+1i

=1=1, or 1+0i1+0i

Feb 10, 2018

See the answer below...

Explanation:

Z=(1+i)/(1-i)+2/(1+i)Z=1+i1i+21+i

=>Z=((1+i)^2+2(1-i))/(1-i^2)" "["simple addition"]Z=(1+i)2+2(1i)1i2 [simple addition]

=>Z=(1+2i+i^2+2-2i)/(1-i^2)Z=1+2i+i2+22i1i2

=>Z=(1+2i-1+2-2i)/(1+1)" "[i^2=-1]Z=1+2i1+22i1+1 [i2=1]

=>Z=1Z=1

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

Hope it helps...
Thank you..

Feb 10, 2018

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