+129840 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
+26388 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:
\(\begin{array}{|rcll|} \hline z &=& x+i\cdot y \quad & | \quad x =\frac{\sqrt{2}} {2} \quad \text{and} \quad y = -\frac{\sqrt{2}} {2} \\ z &=& \frac{\sqrt{2}} {2} - i\cdot \frac{\sqrt{2}} {2} \\ \frac{1}{z} &=& \frac{\frac{\sqrt{2}} {2}}{ (\frac{\sqrt{2}} {2})^2+(-\frac{\sqrt{2}} {2})^2} - i\cdot \frac{-\frac{\sqrt{2}} {2}}{(\frac{\sqrt{2}} {2})^2+(-\frac{\sqrt{2}} {2})^2} \\ \frac{1}{z} &=& \frac{\frac{\sqrt{2}} {2}}{ \frac24+\frac24} - i\cdot \frac{-\frac{\sqrt{2}} {2}}{\frac24+\frac24} \\ \frac{1}{z} &=& \frac{\frac{\sqrt{2}} {2}}{ 1} - i\cdot \frac{-\frac{\sqrt{2}} {2}}{1} \\ \mathbf{\dfrac{1}{z}} &\mathbf{=}& \mathbf{\dfrac{\sqrt{2}} {2} + i\cdot \dfrac{\sqrt{2}} {2} } \\ \hline \end{array}\)
