Complex number. How is this expressed in the form x + iy?

• Jul 12th 2009, 09:20 AM
db5vry
Complex number. How is this expressed in the form x + iy?
The complex number z and its complex conjugate Z satisfy the equation

$\displaystyle 2z + Z = \frac{11 + 7i}{1 + i}$

Find z in the form x + iy. [6]

This is a new topic I am doing in maths and I would like to know how this is done because I don't know the basics in answering it. I would greatly appreciate any help if you could tell me what methods are used. Thanks :)

note: "Z" actually symbolizes a little z with the line on top of it - z's conjugate, to clarify.
• Jul 12th 2009, 09:48 AM
Plato
Quote:

Originally Posted by db5vry
The complex number z and its complex conjugate Z satisfy the equation $\displaystyle 2z + Z = \frac{11 + 7i}{1 + i}$
Find z in the form x + iy. [6]

$\displaystyle \begin{gathered} 2z + \overline z = 3x + yi \hfill \\ \frac{{11 + 7i}} {{1 + i}} = \frac{{\left( {11 + 7i} \right)\left( {1 - i} \right)}} {2} \hfill \\ \end{gathered}$

Now carry on.
• Jul 12th 2009, 09:50 AM
red_dog
First of all $\displaystyle \frac{11+7i}{1+i}=\frac{(11+7i)(1-i)}{2}=9-2i$

$\displaystyle 2(x+yi)+x-yi=9-2i$

$\displaystyle 3x+yi=9-2i$

Then $\displaystyle 3x=9\Rightarrow x=3$

$\displaystyle y=-2$
• Jul 12th 2009, 09:53 AM
mathemanyak
we know z=x+iy and Z=x-iy implies 2x+2iy+x-iy=11+7i/1+i implies 3x+iy=
(11+7i)(1-i)/2 = (11+7+7i-11i)/2 = 9-2i => x=3 and y=-2 => z=3-2i