# Thread: Complex number problem

1. ## Complex number problem

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

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i am getting the answer as

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

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

2. Originally Posted by varunnayudu
Express $\displaystyle z = \frac{\sqrt{2} + \sqrt{6}i}{(\sqrt{3}+i)(\sqrt{2}-\sqrt{6}i)}$ in the form $\displaystyle x+iy$ where $\displaystyle x$ and $\displaystyle y$ are real numbers

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i am getting the answer as

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

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.

3. ## My Work....

My work:

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

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

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

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

4. Hello, varunnayudu!

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

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

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

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

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

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**

I see what he did to the denominator!

. . . This is depressing . . .

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

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

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

5. Originally Posted by varunnayudu
My work:

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

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

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

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