# simple complex conjugates question

• Sep 17th 2008, 05:08 PM
lllll
simple complex conjugates question
show that :
$\displaystyle |(2\overline{z}+5)(\sqrt{2}-i)|=\sqrt{3}|2z+5|$

so for I have:

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

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

now what I don't understand is how $\displaystyle {\color{blue} |(\overline{\sqrt{2}-i})| = |(\overline{\sqrt{2+1}})|} = |\sqrt{3}| = \sqrt{3}$
• Sep 17th 2008, 05:52 PM
Jhevon
Quote:

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

so for I have:

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

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

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

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

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

so, $\displaystyle |\overline{\sqrt{2} - i}| = |\sqrt{2} + i| = \sqrt{(\sqrt{2})^2 + 1^2} = \sqrt{3}$
• Sep 17th 2008, 10:22 PM
Moo
Hello,

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