Complex Number Assignment (1 Viewer)

[alcronin]

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

 

[deswa1]

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

 

[alcronin]

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 :/

 

[alcronin]

Does this mean I can do the same for 2z/ (2+i) + 3-2i = (1-i)Z??

 

[deswa1]

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]

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]

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

 

[alcronin]

Oh crap, I didn't realise they were different signs!! Ok yeah, now I got (1-i)/4. Thanks for spotting that!!

 

[Capt Rifle]

deswa1 is a super hero