Let x = 2e-(p/4)i, y = 3e(p/6)i. Let a + bi = x2 / y3. Find a and b. I do not have any idea for this question.
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Originally Posted by eunse16 Let x = 2e-(p/4)i, y = 3e(p/6)i. Let a + bi = x2 / y3. Find a and b. I do not have any idea for this question. I have no idea what the given numbers are. Are they $\displaystyle x = 2e^{-\frac{p}{4}i}$ $\displaystyle y = 3e^{\frac{p}{6}i}$? And is p pi? 01
Sorry. you are right. I made a mistake. P is pi.
If the numbers are indeed $\displaystyle x=2e^{-\frac{\pi i}{4}}~\&~x=2e^{\frac{\pi i}{6}}$ then $\displaystyle x^2=4e^{-\frac{\pi i}{2}}=-4i$ $\displaystyle y^3=27e^{\frac{\pi i}{2}}=27i$ $\displaystyle \frac{x^2}{y^3}=\frac{-4}{27}$
Using, of course, the laws of exponents: $\displaystyle \left(e^{ax}\right)^n= e^{nax}$ and $\displaystyle \frac{e^{ax}}{e^{bx}}= e^{(a-b)x}$.
Last edited by Isomorphism; May 20th 2009 at 11:08 AM. Reason: Fixed the Latex
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