write this in the exponential polar form

**ilovemymath** · September 25th 2012, 01:04 PM

1)2i/(3e^(4+1))
2) (1-i)/3

**Plato** · September 25th 2012, 02:05 PM

Originally Posted by ilovemymath

1)2i/(3e^(4+1))
2) (1-i)/3

Do you mean $3e^{4+i}~?$

**ilovemymath** · September 25th 2012, 04:18 PM

Yes its 4+i

**Prove It** · September 25th 2012, 04:57 PM

Originally Posted by ilovemymath

1)2i/(3e^(4+1))
2) (1-i)/3

$\displaystyle \begin{align*} \frac{2i}{3e^{4 + i}} &= \frac{2i}{3e^4\,e^i} \\ &= \frac{2}{3e^4}\left( \frac{i}{e^i} \right) \\ &= \frac{2}{3e^4} \left( \frac{e^{\frac{\pi}{2}i}}{e^i} \right) \\ &= \frac{2}{3e^4} \left[ e^{\left(\frac{\pi}{2} - 1\right) i} \right] \end{align*}$

**Prove It** · September 25th 2012, 05:00 PM

Originally Posted by ilovemymath

1)2i/(3e^(4+1))
2) (1-i)/3

$\displaystyle \begin{align*} \frac{1 - i}{3} &= \frac{1}{3} \left( 1 - i \right) \\ &= \frac{1}{3} \left\{ \sqrt{2} \left[ \cos{\left( -\frac{\pi}{4} \right)} + i\sin{\left(-\frac{\pi}{4}\right)} \right] \right\} \\ &= \frac{\sqrt{2}}{3}\,e^{-\frac{\pi}{4}i} \end{align*}$

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