# simple complex conjugates question

• September 17th 2008, 06:08 PM
lllll
simple complex conjugates question
show that :
$|(2\overline{z}+5)(\sqrt{2}-i)|=\sqrt{3}|2z+5|$

so for I have:

$= |(2\overline{z}+5)|\times |(\sqrt{2}-i)|$

$= |(2z+5)| \times |(\overline{\sqrt{2}+i})|$

now what I don't understand is how ${\color{blue} |(\overline{\sqrt{2}-i})| = |(\overline{\sqrt{2+1}})|} = |\sqrt{3}| = \sqrt{3}$
• September 17th 2008, 06:52 PM
Jhevon
Quote:

Originally Posted by lllll
show that :
$|(2\overline{z}+5)(\sqrt{2}-i)|=\sqrt{3}|2z+5|$

so for I have:

$= |(2\overline{z}+5)|\times |(\sqrt{2}-i)|$

$= |(2z+5)| \times |(\overline{\sqrt{2}+i})|$

now what I don't understand is how ${\color{blue} |(\overline{\sqrt{2}-i})| = |(\overline{\sqrt{2+1}})|} = |\sqrt{3}| = \sqrt{3}$

what you wrote is not true. the final answer is, though.

recall, $|x + iy| = \sqrt{x^2 + y^2}$

so, $|\overline{\sqrt{2} - i}| = |\sqrt{2} + i| = \sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{3}$
• September 17th 2008, 11:22 PM
Moo
Hello,

The real equality is $\left|\overline{z}\right|=\left|z\right|$