Question #a53f5

3 Answers
Feb 10, 2018

#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#

Feb 10, 2018

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..

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#