My Math Forum imaginary part of a complex number

 Complex Analysis Complex Analysis Math Forum

 February 20th, 2013, 09:53 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, 11:51 AM #2 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 625 Thanks: 87 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, 12:20 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, 11:37 AM   #4
Math Team

Joined: Aug 2012
From: Sana'a , Yemen

Posts: 1,177
Thanks: 44

Math Focus: Theory of analytic functions
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, 09:21 PM #5 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 625 Thanks: 87 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.

 Tags complex, imaginary, number, part

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post archer18 Complex Analysis 3 March 2nd, 2014 12:46 PM Schmidtacus Complex Analysis 2 February 20th, 2013 09:12 AM skarface Algebra 6 March 10th, 2012 10:01 PM silvetobristol Complex Analysis 0 April 28th, 2010 01:50 AM Malgrif Algebra 5 September 17th, 2009 03:26 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top