Complex number problem

**varunnayudu** · December 29th 2008, 03:05 AM

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

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i am getting the answer as

$x+iy = -\frac{\sqrt{72}}{28} + \frac{\sqrt{104}}{28}i$

is it right or wrong if wrong can u give the right one .

**mr fantastic** · December 29th 2008, 03:16 AM

Originally Posted by varunnayudu

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

----------------------
i am getting the answer as

$x+iy = -\frac{\sqrt{72}}{28} + \frac{\sqrt{104}}{28}i$

is it right or wrong if wrong can u give the right one .

I get i/2.

If you show your working your error(s) can be pointed out.

**varunnayudu** · December 29th 2008, 04:46 AM

My work:

$z = \frac{\sqrt{2}+\sqrt{6}i}{(\sqrt{3}+i)(\sqrt{2}-\sqrt{6}i)}=\frac{\sqrt{2}+\sqrt{6}i}{\sqrt{12}-\sqrt{16}i}$

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

= $ \frac{\sqrt{24}+ \sqrt{32}i + \sqrt{72}i - \sqrt{96}}{(\sqrt{12}-\sqrt{16}i)(\sqrt{12}+ \sqrt{16}i)} $

= $ \frac{- \sqrt{72}+\sqrt{104}i}{12+16} $ = $\frac{- \sqrt{72}+\sqrt{104}i}{28} $

**Soroban** · December 29th 2008, 05:43 AM

Hello, varunnayudu!

My work: . $z \:= \:\frac{\sqrt{2}+\sqrt{6}\,i}{(\sqrt{3}+i)(\sqrt{2 }-\sqrt{6}\,i)}=\frac{\sqrt{2}+\sqrt{6}\,i}{\sqrt{12 }-\sqrt{16}\,i}$ . . . . How? .**

I did some factoring first:

. . $\frac{\sqrt{2} + \sqrt{6}\,i}{(\sqrt{3}+i)(\sqrt{2}-\sqrt{6}\,i)} \;=\;\frac{{\color{red}\rlap{///}}\sqrt{2}\,(1+\sqrt{3}\,i)}{(\sqrt{3}+i){\color{r ed}\rlap{///}}\sqrt{2}\,(1-\sqrt{3}\,i)}$ . $=\;\frac{1+\sqrt{3}\,i}{(\sqrt{3}+i)(1-\sqrt{3}\,i)}$

The denominator is: . $(\sqrt{3}+i)(1-\sqrt{3}\,i) \:=\:\sqrt{3} - 3i + i + \sqrt{3} \:=\:2\sqrt{3} - 2i \:=\:2(\sqrt{3}-i)$

The fraction becomes: . $\frac{1+\sqrt{3}\,i}{2(\sqrt{3}-i)}$

Rationalize: . $\frac{1+\sqrt{3}\,i}{2(\sqrt{3}-i)}\cdot{\color{blue}\frac{\sqrt{3}+i}{\sqrt{3}+i} } \;=\;\frac{\sqrt{3} + i + 3i - \sqrt{3}}{2(3 + 1)} \;=\;\frac{4i}{8} \;=\;\frac{i}{2}$ . . . as promised by mr. f

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

**

I see what he did to the denominator!
. . . This is depressing . . .

$\left(\sqrt{3} + i\right)\left(\sqrt{2} - \sqrt{6}\,i\right) \;=\;\left(\sqrt{3}\right)\left(\sqrt{2}\right) + \left(\sqrt{3}\right)\left(\text{-}\sqrt{6}\,i\right) + (i)\left(\sqrt{2}\right) + (i)\left(\text{-}\sqrt{6}\,i\right)$

. . . . . . . . . . . . . . $= \;\sqrt{6} - \sqrt{18}\,i + \sqrt{2}\,i + \sqrt{6}$

and, of course: . $\begin{array}{ccc}\sqrt{6} + \sqrt{6} &=& \sqrt{12} \\ \\[-4mm] \text{-}\sqrt{18}\,i + \sqrt{2}\,i &=& \text{-}\sqrt{16}\,i \end{array}\quad\hdots\quad NOT\:!$

**mr fantastic** · December 29th 2008, 11:36 AM

Originally Posted by varunnayudu

My work:

$z = \frac{\sqrt{2}+\sqrt{6}i}{(\sqrt{3}+i)(\sqrt{2}-\sqrt{6}i)}=\frac{\sqrt{2}+\sqrt{6}i}{\sqrt{12}-\sqrt{16}i}$ Mr F says: The first mistake is in the first line. This denominator is not correct. Try expanding again.

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

= $ \frac{\sqrt{24}+ \sqrt{32}i + \sqrt{72}i - \sqrt{96}}{(\sqrt{12}-\sqrt{16}i)(\sqrt{12}+ \sqrt{16}i)} $

= $ \frac{- \sqrt{72}+\sqrt{104}i}{12+16} $ = $\frac{- \sqrt{72}+\sqrt{104}i}{28} $

..

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