Complex number question help please :D (1 Viewer)

[macrazy]

Solve for z and express in the form a + ib:
(2z)/(2+i) + 3 - 2i = (1-i)z
I can't seem to isolate z, it's probably something really simple that I'm missing but it's killing me (only had one lesson of ext. 2 so far and already questioning my decision)

 

[iStudent]

2z/2+i can be written as z (2/(2+i)). Does that help? And yes, it is something very simple.

 

[itsalberttrinh]

Clean up the 2z/2+i first by multiplying by its conjugate. Then basically rearrange making Z the subject.
2z(2-i)/5 + (3-2i) = (1-i)z
2z(2-i) + (15-10i) = 5z(1-i)
z = (15-10i)/(1-3i) [Times by conjugate]
z = (45+35i)/10
z = 9/2 + 7i/2

 

[faisalabdul16]

<a href="http://www.codecogs.com/eqnedit.php?latex=\frac{2z}{2&plus;i}&plus;&space;3-2i&space;=&space;z(1-i)\\&space;\\&space;3-2i&space;=&space;z(1-i&space;-\frac{2}{2&plus;i})\\&space;\\&space;$Rearrange&space;and&space;solve&space;for&space;z" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{2z}{2&plus;i}&plus;&space;3-2i&space;=&space;z(1-i)\\&space;\\&space;3-2i&space;=&space;z(1-i&space;-\frac{2}{2&plus;i})\\&space;\\&space;$Rearrange&space;and&space;solve&space;for&space;z" title="\frac{2z}{2+i}+ 3-2i = z(1-i)\\ \\ 3-2i = z(1-i -\frac{2}{2+i})\\ \\ $Rearrange and solve for z" /></a>