# Thread: Complex Number problem

1. ## Complex Number problem

Question: Express $z = \ \frac{\sqrt{2} + \sqrt{6} i}{(\sqrt{3} + i)(\sqrt{2} - \sqrt{6}i)}$ in the form x+iy where x and y are real numbers

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

= $\frac{\sqrt{2} + \sqrt{6} i}{(\sqrt{3} + i)(\sqrt{2} - \sqrt{6}i)}$ = ................= $\frac{\sqrt{104}i - \sqrt{72}}{24}$......Is this correct???

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

3. ## This is what i have done?

Originally Posted by pickslides
To do this problem you need to expand the denominator to the form $x+yi$ after this you need to mulitply the both numerator and denominator by the complex conjugate $x-yi$ . Is this what you did?
$\frac{\sqrt{2} + \sqrt{6}i}{(\sqrt{3}+i)(\sqrt{2}-\sqrt{6}i)}$

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

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

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

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

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

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

= $\frac{ \sqrt{104}i - \sqrt{72}}{28}$.............Could any body please tell me what i have done wrong and how should i rectify it ???

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

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

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

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

. . .

5. Originally Posted by zorro
$$= $\frac{\sqrt{2} + \sqrt{6}i}{\sqrt{6}- \sqrt{18}i + \sqrt{2}i -\sqrt{6}i^2}$ = $\frac{\sqrt{2} + \sqrt{6}i}{\sqrt{6}- \sqrt{16}i + \sqrt{6}}$ $- \sqrt{18}i + \sqrt{2}i \neq - \sqrt{16}i$ $- \sqrt{18}i + \sqrt{2}i = - \sqrt{9\times 2}i + \sqrt{2}i =- 3\sqrt{ 2}i + \sqrt{2}i =-2\sqrt{ 2}i$ also $\sqrt{6}+\sqrt{6} = 2\sqrt{6}\neq \sqrt{12}$ 6. Originally Posted by dedust = $\frac{\sqrt{2} + \sqrt{6}i}{\sqrt{6}- \sqrt{18}i + \sqrt{2}i -\sqrt{6}i^2}$ = $\frac{\sqrt{2} + \sqrt{6}i}{2\sqrt{6}- \sqrt{8}i}$$$

. . .
This can be simplified further $- \sqrt{8}i = - 2\sqrt{2}i$

7. ## Thanks Pickslide and Mark Goodstein

Thanks u for helping me
pickslides and Mark Goodstein

cheers

8. ## Is this correct?

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

cheers

The result which i am getting is $- \frac{1}{2}$

9. ## Is the answer correct ?

Originally Posted by pickslides
This can be simplified further $- \sqrt{8}i = - 2\sqrt{2}i$

Mite it the answer correct?