Tried getting a common denominator for both fractions

[aisha]

find real and complex part of z: z/z+2=2-i

I can't factor out the z because of the 2 in the denominator. The i can be written as the square root of -1 but that doesn't help. I tried multiplying by the conjugate to get z alone but nope not any good. I am doing something wrong can someone please start me off ?

Right now i tried getting a common denominator for both fractions (2z) and then I got stuck at 3z/2z=2-i I think this is wrong.


Help Me Please

 

[ceptimus]

Try multiplying both sides by (z + 2)

 

[Gokul43201]

Let z = a + ib, cross-multiply and compare real and imaginary parts.

 

[aisha]

Let z = a + ib, cross-multiply and compare real and imaginary parts.

what am i cross multiplying ?

I tried multiplying both sides by (z+2) that does not work because then there will be two variable i and z on the right hand side.

 

[Gokul43201]

'i' is not a variable. As, you mentioned before, it is the square root of -1. Do you understand what real and imaginary parts mean ?

If you have an equation of the form a + ib = c + id, then a = c and b = d.

 

[aisha]

yes I understand. Well this is what I did, l multiplied both sides by (z+2) this gave me z alone on the left hand side and on the right I got z=(2-i)(z+2) I expanded this and got
z=2z+4-zi-2i now i don't know what to do

The solution is z=3+i

 

[nolachrymose]

Try to collect all terms "containing" z onto one size of the equation, and factoring.

 

[Hurkyl]

You seem to be having a little trouble both with algebra and complex numbers... it might help to try working on similar problems that only involve real numbers. For instance, solve x/(x+2) = 2 for x.

 

[aisha]

You seem to be having a little trouble both with algebra and complex numbers... it might help to try working on similar problems that only involve real numbers. For instance, solve x/(x+2) = 2 for x.

That I can do x=2x+4 but this question is just hurting my head

 

[Gokul43201]

That I can do x=2x+4

No...that's not a solution. A solution would look like "x = blah", where 'blah' is a number that does not involve x. Can you get that from x =2x + 4 ?

 

[BobG]

yes I understand. Well this is what I did, l multiplied both sides by (z+2) this gave me z alone on the left hand side and on the right I got z=(2-i)(z+2) I expanded this and got
z=2z+4-zi-2i now i don't know what to do

Try to collect all terms "containing" z onto one size of the equation, and factoring.

Did you try nolachrymose's suggestion. Once you have all the z's on the same side of the equation, you can factor out the z, leaving you with

z(some numbers)

Now, if you take and divide both sides of the equation by (some numbers), you can solve the right hand side and do wind up with 3+i, as you said.