Algebra

1  /    ((sqrt 2 -isqrt2) / 2)   =  2 / (sqrt2-isqrt2)

2 / [ √2 - √2i]   =

2 / [ √2 ( 1 - i) ]  =

√2 / [ 1 -  i ]    =

√2 [ 1 + i ]  / [ (1 - i) ( 1 + i ) ] =

√2 [ 1 + i ]  / [ 1 - i^2]  =

√2 [ 1 + i ] / 2  =

[ 1 + i ] / √2

Find the reciprocal of

$\frac{\sqrt{2}} {2} - i\cdot \frac{\sqrt{2}} {2}$    

( The letter "i" refers to the complex number)

Formula:

$\begin{array}{|rcll|} \hline z &=& x+i\cdot y \\ \frac{1}{z} &=& \frac{x}{x^2+y^2} - i\cdot \frac{y}{x^2+y^2} \\ \hline \end{array}$

We have:

$x =\frac{\sqrt{2}} {2} \quad \text{and} \quad y = -\frac{\sqrt{2}} {2}$

so the reciprocal is:

$\begin{array}{|rcll|} \hline z &=& x+i\cdot y \quad & | \quad x =\frac{\sqrt{2}} {2} \quad \text{and} \quad y = -\frac{\sqrt{2}} {2} \\ z &=& \frac{\sqrt{2}} {2} - i\cdot \frac{\sqrt{2}} {2} \\ \frac{1}{z} &=& \frac{\frac{\sqrt{2}} {2}}{ (\frac{\sqrt{2}} {2})^2+(-\frac{\sqrt{2}} {2})^2} - i\cdot \frac{-\frac{\sqrt{2}} {2}}{(\frac{\sqrt{2}} {2})^2+(-\frac{\sqrt{2}} {2})^2} \\ \frac{1}{z} &=& \frac{\frac{\sqrt{2}} {2}}{ \frac24+\frac24} - i\cdot \frac{-\frac{\sqrt{2}} {2}}{\frac24+\frac24} \\ \frac{1}{z} &=& \frac{\frac{\sqrt{2}} {2}}{ 1} - i\cdot \frac{-\frac{\sqrt{2}} {2}}{1} \\ \mathbf{\dfrac{1}{z}} &\mathbf{=}& \mathbf{\dfrac{\sqrt{2}} {2} + i\cdot \dfrac{\sqrt{2}} {2} } \\ \hline \end{array}$