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

• July 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

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

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.
• July 12th 2009, 09:48 AM
Plato
Quote:

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

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

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

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

$3x+yi=9-2i$

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

$y=-2$
• July 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