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 February 20th, 2013, 10:53 AM #1 Senior Member   Joined: Sep 2010 From: Germany Posts: 153 Thanks: 0 imaginary part of a complex number what is the imaginary part of $z=(i-1)^{i}$
 February 20th, 2013, 12:51 PM #2 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 633 Thanks: 94 Math Focus: Electrical Engineering Applications Re: imaginary part of a complex number $z=(i-1)^i$ $z=\left(sqrt{2} \cdot e^{i \frac{3 \pi}{4}}\right)^i$ $z=(\sqrt{2})^i \cdot e^{i \cdot i \frac{3 \pi}{4}}=(\sqrt{2})^i \cdot e^{-\frac{3 \pi}{4}}$ Since $\ \ln{((\sqrt{2})^i})=i \cdot \ln{(\sqrt{2})}$ $z=e^{i \cdot \ln{(\sqrt{2})}} \cdot e^{-\frac{3 \pi}{4}}$ $z=e^{-\frac{3 \pi}{4}} \cdot e^{i \cdot \ln{(\sqrt{2})}}=e^{-\frac{3 \pi}{4}}\left(\cos{(\ln({\sqrt{2})})}+i \cdot \sin{(\ln{(\sqrt{2})})}\right)$ $z \approx 0.089145+0.032195i$
 February 20th, 2013, 01:20 PM #3 Senior Member   Joined: Sep 2010 From: Germany Posts: 153 Thanks: 0 Re: imaginary part of a complex number thank you !
February 21st, 2013, 12:37 PM   #4
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Re: imaginary part of a complex number

Quote:
 Originally Posted by jks Since $\ \ln{((\sqrt{2})^i})=i \cdot \ln{(\sqrt{2})}$
You have to justify this ...

 February 22nd, 2013, 10:21 PM #5 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 633 Thanks: 94 Math Focus: Electrical Engineering Applications Re: imaginary part of a complex number I'm glad that you asked for justification as it gave me cause to study complex exponentiation and complex logarithms further. If the justification that I give below is inadequate, then I will require, and appreciate, help to make it correct and complete. I think that you might have two main objections. The first may be that as this reference points out, for a real number, b, to complex powers u and z, then in general, $\ (b^z)^u=b^{zu} \$ is not valid (see the sub-sub section "Complex exponents with positive real bases" down below the sub section with the same title; I could not figure out how to link there directly since the titles are the same). The second objection may be that I used the principal value of $\ \ln{\sqrt{2}} \$ but I did not specifically state that I was doing so. Further down on the same reference, a complex number, w, to a complex power, z is defined as $w^z=e^{z\ln w \$. Also, this reference gives the logarithm of the principal value of a complex number in polar form, $\ z=r\cdot e^{i \theta} \$ as: $\ln{z}=\ln{r}+i \theta \$ (which I assume extends to any value of $\ \theta \$). Using these definitions, then: $(\sqrt{2})^i=\left(\sqrt{2} \cdot e^{i2\pi k}\right)^i \$ with $\ k \in \mathbb{Z} \$. Taking the principal value with $\ k=0 \$ gives $\ (\sqrt{2})^i \$, and from the definition given above: $(\sqrt{2})^i=e^{i \ln{\sqrt{2}} \$ (which is the next step after the 'Since ...' in my derivation above). Of course, this is equivalent to: $(\sqrt{2})^i=1e^{i \ln{\sqrt{2}} \$ and if we take the log, we get: $\ln{(\sqrt{2})^i}=\ln{(1e^{i \ln{\sqrt{2}}})}=\ln{1}+i\ln{\sqrt{2}}=0+i\ln{\sqr t{2}}=i\ln{\sqrt{2}}$ Just for grins, taking a non principal value: $\left(\sqrt{2} \cdot e^{i2\pi}\right)^i=e^{i(\ln{\sqrt{2}}+2\pi i)}=e^{-2\pi}e^{i \ln{\sqrt{2}}}$ $\left(\sqrt{2} \cdot e^{i2\pi}\right)^i=e^{-2\pi}\cos{(\ln{\sqrt{2}})}+ie^{-2\pi}\sin{(\ln{\sqrt{2}})}$ I hope that this is correct, and that the justification is adequate.

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