My Math Forum How to deal with powers of i

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 May 5th, 2012, 06:56 PM #1 Newbie   Joined: May 2012 Posts: 1 Thanks: 0 How to deal with powers of i While revising for my exam this week I came across a final problem in the set; Express (-1+i)^i in the form a+ib I have gotten until 2^(i/2).e^(-3pi/4) However what should I do from here in order to bring the i down into standard form
 May 5th, 2012, 08:53 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: How to deal with powers of i Use the identity: $x=e^{\ln(x)}$ and Euler's formula to get: $e^{-\frac{3\pi}{4}}\cdot e^{\ln$$2^{\frac{i}{2}}$$}=e^{-\frac{3\pi}{4}}$$e^{\frac{\ln(2)}{2}i}$$=e^{-\frac{3\pi}{4}}$$\cos\(\frac{\ln(2)}{2}$$+i\sin$$\ frac{\ln(2)}{2}$$\)=$ $e^{-\frac{3\pi}{4}}\cos$$\frac{\ln(2)}{2}$$+ie^{-\frac{3\pi}{4}}\sin$$\frac{\ln(2)}{2}$$$
 May 5th, 2012, 10:36 PM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond Re: How to deal with powers of i \begin{align*}(-1\,+\,i)^i\,&=\,a\,+\,bi \\ i\ln(-1\,+\,i)\,&=\,\ln$$\sqrt{a^2\,+\,b^2}$$\,+\,i\thet a \\ -\frac{3\pi}{4}\,+\,i\ln(\sqrt{2})\,&=\,\ln$$\sqrt{ a^2\,+\,b^2}$$\,+\,i\theta \\ \tan$$\ln(\sqrt{2})$$\,&=\,\tan(\theta)\,=\,\frac{ b}{a} \\ b\,&=\,a\,\cdot\,\tan$$\ln(\sqrt{2})$$ \\ e^{-3\pi/4}\,&=\,\sqrt{a^2\,+\,a^2\tan^2(\ln(\sqrt{2}))} \\ &=\,a\,\cdot\,\sec(\ln(\sqrt{2})) \\ a\,&=\,\frac{\cos$$\ln(\sqrt{2})$$}{e^{3\pi/4}} \\ b\,&=\,\frac{\sin(\ln(\sqrt{2}))}{e^{3\pi/4}}\end{align*} $(-1\,+\,i)^i\,=\,\frac{\cos$$\ln(\sqrt{2})$$}{e^{3\p i/4}}\,+\,i\frac{\sin(\ln(\sqrt{2}))}{e^{3\pi/4}}$
 May 9th, 2012, 03:53 AM #4 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,963 Thanks: 1148 Math Focus: Elementary mathematics and beyond Re: How to deal with powers of i Here's a general solution to a complex number raised to a complex number: $(a\,+\,bi)^{c\,+\,di}\,=\,e^{$$c\ln\sqrt{a^2\,+\,b ^2}\,-\,d\theta$$}\,\cdot\,\cos$$c\theta\,+\,d\ln\sqrt{a ^2\,+\,b^2}$$\,+\,ie^{$$c\ln\sqrt{a^2\,+\,b^2}\,-\,d\theta$$}\,\cdot\,\sin$$c\theta\,+\,d\ln\sqrt{a ^2\,+\,b^2}$$$ Computing $\theta$, the angle between the complex vector and the positive real axis, in radians, using the signs of $a$ and $b$: $\begin{tabular}{ c c c c} a & b & Quadrant \\ + & + & 1 \\ - & + & 2 \\ - & - & 3 \\ + & - & 4 \end{tabular}$ $\text{Q1: use }\theta\,=\,\tan^{\small{-1}}$$\frac{b}{a}$$$ $\text{Q2: use }\theta\,=\,\tan^{\small{-1}}$$\frac{b}{a}$$\,+\,\pi$ $\text{Q3: use }\theta\,=\,\tan^{\small{-1}}$$\frac{b}{a}$$\,-\,\pi$ $\text{Q4: use }\theta\,=\,\tan^{\small{-1}}$$\frac{b}{a}$$$

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