Complex Number Assignment (1 Viewer)

deswa1

Well-Known Member
Joined
Jul 12, 2011
Messages
2,258
Gender
Male
HSC
2012
Does anyone know how to solve this, I'm having great difficulty:

Solve for Z:

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

Any help would be appreciated, thanks :)
I'll just give a few pointers and if you still need more help, then I'll write more:

1. Put the RHS all on the same denominator
2. Take inverses of both sides
3. Realise the denominator
4. Finished
 

deswa1

Well-Known Member
Joined
Jul 12, 2011
Messages
2,258
Gender
Male
HSC
2012
I got that Z = 1/2i I hope that's right :) Thank you so much, I was going towards substituting in Z = x + iy and everything, was getting messy :/
I didn't get that- I just did it very quickly so it might be wrong but I got z=(1-i)/4. Care to post your working?
 

alcronin

New Member
Joined
Jul 25, 2011
Messages
20
Location
Coffs Harbour, New South Wales, Australia
Gender
Female
HSC
2013
1/Z = ((1+i)^2 + 2)/(1-i)
= (1+2i-1+2)/(1-i)
= (2i+2)/(1-i)
= (1+2i)/(1-i) X (1+i)/(1+i)
= (2+2i+2i-2)/(1+1)
= 4i/2
= 2i
therefore, Z = 1/2i


Sorry for it being hard to understand, they don't have equation editors on here :)
 

deswa1

Well-Known Member
Joined
Jul 12, 2011
Messages
2,258
Gender
Male
HSC
2012
Why is your first line:

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

It should be:

1/Z= ((1+i)(1-i) +2) /(1-i) because you are putting everything on top of 1-i
 

Users Who Are Viewing This Thread (Users: 0, Guests: 1)

Top