Question

Find the polar form of 1+ i / 1- i ​

Answer 1

hmmm let's start off by using the conjugate of the denominator to rationalize the denominator.

`\bf \cfrac{1+i}{1-i}\implies \cfrac{1+i}{1-i}\cdot \cfrac{1+i}{1+i}\implies \cfrac{(1+i)^2}{\underset{\textit{difference of squares}}{(1-i)(1+i)}}\implies \cfrac{\stackrel{FOIL}{1^2+2i+i^2}}{1^2-i^2} \\\\\\ \stackrel{\textit{recalling that }i^2=-1}{\cfrac{1+2i+(-1)}{1-(-1)}}\implies \cfrac{2i}{1+1}\implies \cfrac{~~\begin{matrix} 2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~i}{~~\begin{matrix} 2 \\[-0.7em]\cline{1-1}\\[-5pt]\end{matrix}~~}\implies i\implies \stackrel{\textit{a+bi form}}{0 + 1i}`

so the equation is really "i" in disguise, now, our coordinates are 0 and 1 for that complex point, we can't quite use tan⁻¹ to get the angle, since denominator is 0, meaning the point is lying on one of the axis, well, the rectangular (0,1) is right up above on the y-axis, namely at π/2.

`\bf \begin{cases} r = √(0^2+1^2)\\ \qquad 1\\ \theta =(\pi )/(2) \end{cases} \qquad \qquad 1\left[cos\left( (\pi )/(2) \right) +i~sin\left( (\pi )/(2) \right) \right]`