write this in the exponential polar form
1)2i/(3e^(4+1))
2) (1-i)/3
Re: write this in the exponential polar form
Quote:
Originally Posted by
ilovemymath 
1)2i/(3e^(4+1))
2) (1-i)/3
Do you mean 
Re: write this in the exponential polar form
Re: write this in the exponential polar form
Quote:
Originally Posted by
ilovemymath 1)2i/(3e^(4+1))
2) (1-i)/3
![\displaystyle \begin{align*} \frac{2i}{3e^{4 + i}} &= \frac{2i}{3e^4\,e^i} \\ &= \frac{2}{3e^4}\left( \frac{i}{e^i} \right) \\ &= \frac{2}{3e^4} \left( \frac{e^{\frac{\pi}{2}i}}{e^i} \right) \\ &= \frac{2}{3e^4} \left[ e^{\left(\frac{\pi}{2} - 1\right) i} \right] \end{align*}](http://latex.codecogs.com/png.latex?\displaystyle \begin{align*} \frac{2i}{3e^{4 + i}} &= \frac{2i}{3e^4\,e^i} \\ &= \frac{2}{3e^4}\left( \frac{i}{e^i} \right) \\ &= \frac{2}{3e^4} \left( \frac{e^{\frac{\pi}{2}i}}{e^i} \right) \\ &= \frac{2}{3e^4} \left[ e^{\left(\frac{\pi}{2} - 1\right) i} \right] \end{align*})
Re: write this in the exponential polar form
Quote:
Originally Posted by
ilovemymath 1)2i/(3e^(4+1))
2) (1-i)/3
![\displaystyle \begin{align*} \frac{1 - i}{3} &= \frac{1}{3} \left( 1 - i \right) \\ &= \frac{1}{3} \left\{ \sqrt{2} \left[ \cos{\left( -\frac{\pi}{4} \right)} + i\sin{\left(-\frac{\pi}{4}\right)} \right] \right\} \\ &= \frac{\sqrt{2}}{3}\,e^{-\frac{\pi}{4}i} \end{align*}](http://latex.codecogs.com/png.latex?\displaystyle \begin{align*} \frac{1 - i}{3} &= \frac{1}{3} \left( 1 - i \right) \\ &= \frac{1}{3} \left\{ \sqrt{2} \left[ \cos{\left( -\frac{\pi}{4} \right)} + i\sin{\left(-\frac{\pi}{4}\right)} \right] \right\} \\ &= \frac{\sqrt{2}}{3}\,e^{-\frac{\pi}{4}i} \end{align*})
