# Complex Number problem

• Dec 14th 2009, 07:58 PM
zorro
Complex Number problem
Question: Express $\displaystyle z = \ \frac{\sqrt{2} + \sqrt{6} i}{(\sqrt{3} + i)(\sqrt{2} - \sqrt{6}i)}$ in the form x+iy where x and y are real numbers

Solution :
$\displaystyle z = \ \frac{\sqrt{2} + \sqrt{6} i}{(\sqrt{3} + i)(\sqrt{2} - \sqrt{6}i)}$

= $\displaystyle \frac{\sqrt{2} + \sqrt{6} i}{(\sqrt{3} + i)(\sqrt{2} - \sqrt{6}i)}$ = ................= $\displaystyle \frac{\sqrt{104}i - \sqrt{72}}{24}$......Is this correct???
• Dec 14th 2009, 08:03 PM
pickslides
To do this problem you need to expand the denominator to the form $\displaystyle x+yi$ after this you need to mulitply the both numerator and denominator by the complex conjugate $\displaystyle x-yi$ . Is this what you did?
• Dec 14th 2009, 08:20 PM
zorro
This is what i have done?
Quote:

Originally Posted by pickslides
To do this problem you need to expand the denominator to the form $\displaystyle x+yi$ after this you need to mulitply the both numerator and denominator by the complex conjugate $\displaystyle x-yi$ . Is this what you did?

$\displaystyle \frac{\sqrt{2} + \sqrt{6}i}{(\sqrt{3}+i)(\sqrt{2}-\sqrt{6}i)}$

=$\displaystyle \frac{\sqrt{2} + \sqrt{6}i}{\sqrt{6}- \sqrt{18}i + \sqrt{2}i -\sqrt{6}i^2}$

=$\displaystyle \frac{\sqrt{2} + \sqrt{6}i}{\sqrt{6}- \sqrt{16}i + \sqrt{6}}$

=$\displaystyle \frac{\sqrt{2} + \sqrt{6}i}{\sqrt{12}- \sqrt{16}i}$

= $\displaystyle \frac{\sqrt{2} + \sqrt{6}i}{\sqrt{12}- \sqrt{16}i} \times \frac{\sqrt{12}+ \sqrt{16}i}{\sqrt{12}+ \sqrt{16}i}$

= $\displaystyle \frac{\sqrt{24} + \sqrt{32}i + \sqrt{72}i + \sqrt{96}i^2}{12 + \sqrt{192}i - \sqrt{192}i - 16 i^2}$

= $\displaystyle \frac{\sqrt{24} + \sqrt{104}i - \sqrt{96}}{12 + 16}$

= $\displaystyle \frac{ \sqrt{104}i - \sqrt{72}}{28}$.............Could any body please tell me what i have done wrong and how should i rectify it ???
• Dec 14th 2009, 09:11 PM
dedust
$\displaystyle \frac{\sqrt{2} + \sqrt{6}i}{(\sqrt{3}+i)(\sqrt{2}-\sqrt{6}i)}$

=$\displaystyle \frac{\sqrt{2} + \sqrt{6}i}{\sqrt{6}- \sqrt{18}i + \sqrt{2}i -\sqrt{6}i^2}$

=$\displaystyle \frac{\sqrt{2} + \sqrt{6}i}{2\sqrt{6}- \sqrt{8}i}$

= $\displaystyle \frac{\sqrt{2} + \sqrt{6}i}{2\sqrt{6}- \sqrt{8}i} \times \frac{2\sqrt{6} + \sqrt{8}i}{2\sqrt{6} + \sqrt{8}i}$

. . .
• Dec 15th 2009, 11:57 AM
pickslides
Quote:

Originally Posted by zorro
$$=\displaystyle \frac{\sqrt{2} + \sqrt{6}i}{\sqrt{6}- \sqrt{18}i + \sqrt{2}i -\sqrt{6}i^2} =\displaystyle \frac{\sqrt{2} + \sqrt{6}i}{\sqrt{6}- \sqrt{16}i + \sqrt{6}} \displaystyle - \sqrt{18}i + \sqrt{2}i \neq - \sqrt{16}i \displaystyle - \sqrt{18}i + \sqrt{2}i = - \sqrt{9\times 2}i + \sqrt{2}i =- 3\sqrt{ 2}i + \sqrt{2}i =-2\sqrt{ 2}i  also \displaystyle \sqrt{6}+\sqrt{6} = 2\sqrt{6}\neq \sqrt{12} • Dec 15th 2009, 11:59 AM pickslides Quote: Originally Posted by dedust =\displaystyle \frac{\sqrt{2} + \sqrt{6}i}{\sqrt{6}- \sqrt{18}i + \sqrt{2}i -\sqrt{6}i^2} =\displaystyle \frac{\sqrt{2} + \sqrt{6}i}{2\sqrt{6}- \sqrt{8}i}$$

. . .

This can be simplified further $\displaystyle - \sqrt{8}i = - 2\sqrt{2}i$
• Dec 20th 2009, 03:40 PM
zorro
Thanks Pickslide and Mark Goodstein
Thanks u for helping me
pickslides and Mark Goodstein

cheers (Beer)
• Dec 20th 2009, 08:45 PM
zorro
Is this correct?
Quote:

Originally Posted by zorro
Thanks u for helping me
pickslides and Mark Goodstein

cheers (Beer)

The result which i am getting is $\displaystyle - \frac{1}{2}$
• Dec 20th 2009, 08:46 PM
zorro
This can be simplified further $\displaystyle - \sqrt{8}i = - 2\sqrt{2}i$