Complex ExponentiationDate: 02/20/2015 at 19:27:47 From: Jacob Subject: Raising complex numbers to/and complex powers Okay, I have a "biggie" for you. I am trying (and failing) to raise a complex number to a complex power. Actually, the different cases involving complex powers with and without base-e have been scattered across your archives, so I was thinking I'd make it easier for the next person to find an answer all in one place. (Also, I'm not really sure I'm doing half of them right.) Would you mind explaining all four cases? I know Euler's Identity, (e^(xi) = cos(x) + isin(x)), and the natural logarithm identity a = e^(ln(a)). But I have been looking for MANY hours on how to do this (mainly the last one), and I cannot find a decent explanation of how to do it by hand (well, not COMPLETELY by hand...). But mostly, I'm having trouble solving x^y when BOTH x and y are complex numbers. Okay, here are my four example cases. I DO think that I got this one right: e^(3i) = cos(3) + isin(3) = .989992 + .141120i I think that I got this one right, too: e^(2 + 3i) = e^2*e^(3i) = e^2*cos(3) + isin(3) = -7.315110 + .141120i I'm not sure if I got this one right: 6^(2 + 3i) = e^(ln(6^(2 + 3i))) = e^((2 + 3i)ln(6)) = e^(2ln(6))*e^(3i) = 36*cos(3) + isin(3) = -35.639729 + .141120i Here's the one that I really can't seem to figure out: (2 + 3i)^(3 + 2i) = e^(ln((2 + 3i)^(3 + 2i))) = e^((3 + 2i)ln(2 + 3i)) = e^((3 + 2i)^(ln(2 + 3i))) ... And that's where I can't continue. I know you can't split it into its respective parts, because you're raising the WHOLE base (in this case, 3 + 2i) to the power (in this one, ln(2 + 3i)). The most difficult part about these, I think, is the fact that you're breaking apart everything you've learned. I guess we have Euler to thank for making it even possible; but multiplying a number by itself a non-real number of times already strains the intuition. Can you help me out here? For the life of me, I cannot understand this one. I'd appreciate it forever. I'm one of those people who can't sleep until they figure out the problem that they've started. Any help at all is appreciated! Thank you!! Date: 02/20/2015 at 22:14:48 From: Doctor Ali Subject: Re: Raising complex numbers to/and complex powers Hi Jacob! Thanks for writing to Dr. Math. I think the following answer from the Dr. Math archives will tell you what you want to know: http://mathforum.org/library/drmath/view/52251.html I found it by searching on these keywords: complex power You can find other similar answers by doing the same search at http://mathforum.org/library/drmath/mathgrepform.html I hope this helps. Let me know if you'd like to talk about this some more, or if you have any other questions. Please write back if you still have any difficulties. - Doctor Ali, The Math Forum http://mathforum.org/dr.math/ Date: 02/22/2015 at 16:38:20 From: Jacob Subject: Raising complex numbers to/and complex powers Dear Dr. Ali, Thank you so much for your help! Unfortunately, I still don't completely understand a complex power raised to a complex power.... Would it be possible to explain it a little more in depth for me? Thank you!! Jacob Date: 02/23/2015 at 23:30:34 From: Doctor Ali Subject: Re: Raising complex numbers to/and complex powers Hi Jacob! Thanks for writing to Dr. Math. Let's say that we want to find z1^z2, where z1 = a + b i z2 = c + d i This is the most general case in raising complex numbers to complex powers. We can write it as (a + bi)^(c + di) = (a + bi)^c x (a + bi)^(di) The first part, (a + bi)^c, is pretty easy. If c is an integer, you'll just need to use the binomial expansion and expand (a + bi)^c and do the calculation, bearing in mind that i^2 = -1. If it's not an integer, you'll need to write (a + bi) in polar form and use De Moivre's Formula to find the answer in polar form. That is, a + bi = r exp(i t) So, (a + bi)^c = (r exp(i t))^c = r^c exp(i t c) It seems you don't have a problem with the first part, but in case you don't know about this, please take a look at the pages below: http://mathforum.org/library/drmath/sets/select/dm_demoivre.html http://en.wikipedia.org/wiki/De_Moivre's_formula Now for the second part, we want to exaluate (a + bi)^(di). We can write it as (a + bi)^(di) = ((a + bi)^d)^i The term (a + bi)^d is just like above. Now, you have a complex number that you want to raise to i-th power. You can easily do this using the same method discussed in the below page from our archive: http://mathforum.org/library/drmath/view/53860.html I will let you do this part yourself as an exercise. What you are supposed to do is to find M and N in the expression below: (x + yi)^i = M + N i Once you master this technique, you'll see that you can directly separate the real part and imaginary part of the power in the original question and use De Moivre's Formula to find the result of each. Please write back if you still have any difficulties. - Doctor Ali, The Math Forum http://mathforum.org/dr.math/ Date: 02/26/2015 at 10:25:09 From: Jacob Subject: Raising complex numbers to/and complex powers Hi Dr. Ali, I apologize for the late reply. I've had a lot going on. I did have a chance to check out the links, and they explain a lot to me. However, I still am not sure how to get past a certain part in solving (2 + 3i)^(3 + 2i) I can get this far: -46 + 9i(-5 + 12i)^i But from there, I have no idea where to go. I didn't see that they did it in those other answered questions, either. I know you can't "distribute" the exponent (in this case, i), and I don't think you can make a logarithm base -5 + 12i of (x/(-46 - 9i)). And I obviously can't write out and multiply -5 + 12i an i number of times.... This is driving me crazy! Thanks for all of your help! Jacob PS. I saw in another article that you can take the i index of a number (basically, raise it to a (1/i) power). I don't really understand how they got the answers that they did. Can you explain that to me (or send me a link with the work written out)? I'm so sorry, I've created more questions for you. Thank you so much!! Date: 02/26/2015 at 23:05:32 From: Doctor Peterson Subject: Re: Raising complex numbers to/and complex powers Hi, Jacob. Here is what I would do to find (2 + 3i)^(3 + 2i). We first have to express 2 + 3i in polar form. We find that the angle (argument) is arctan(3/2), which I will just call A for now to simplify my work, and the length (modulus) is sqrt(2^2 + 3^2) = sqrt(13). Thus, 2 + 3i = sqrt(13))*e^(Ai) = e^(ln(sqrt(13)) + Ai) Let's use R to represent ln(sqrt(13)), so 2 + 3i = e^(R + Ai) Now, (2 + 3i)^(3 + 2i) = (e^(R + Ai))^(3 + 2i) = e^((R + Ai)(3 + 2i)) = e^((3R-2A) + (2R + 3A)i) = e^(3R-2A)*e^(2R + 3A)i) = e^(3R-2A)[cos(2R + 3A) + sin(2R + 3A)i] = e^(3R-2A)cos(2R + 3A) + e^(3R-2A)sin(2R + 3A)i Numerically, A = arctan(3/2) = 0.98279 R = ln(sqrt(13)) = 1.28247 3R - 2A = 1.88183 e^(3R - 2A) = 6.5655 2R + 3A = 5.51333 cos(2R + 3A) = 0.718012 sin(2R + 3A) = -0.6960310 So our result is 6.5655*0.718012 + 6.5655*-0.6960310i = 4.714 - 4.569i Does that give you a better sense of where an answer like this comes from? - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 03/08/2015 at 10:05:43 From: Jacob Subject: Thank you (Raising complex numbers to/and complex powers) Dr. Ali, Again, sorry for the late response. I get it now. Thank you so much! This is awesome!! I really appreciate all that you have done for me. It means so much. I wish you the best! Happy Math! Jacob |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/