# Complex Powers

by Lonewolf
Tags: complex, powers
 P: 333 How do we express complex powers of numbers (e.g. 21+i) in the form a+bi, or some other standard form of representation for complex numbers?
 Math Emeritus Sci Advisor Thanks PF Gold P: 39,310 First, of course, 21+i= 2*2i so the question is really about 2i (or, more generally, abi). Specfically, look at eix. It is possible to show (using Taylor's series) that e^(ix)= cos(x)+ i sin(x). a^(bi)= e^(ln(a^(bi))= e^(bi*ln(a))= cos(b ln(a))+ i sin(b ln(a)) = cos(ln(a^b))+ i sin(ln(a^b)) For your particular case, 2^i= cos(ln(2))+ i sin(ln(2)) = 0.769+ 0.639 i. 2^(1+i)= 2(0.769+ 0.639i)= 0.1538+ 1.278 i.
P: 333
 21+i= 2*2i
Now why didn't I see that? Oh well, thanks for pointing it out.

 Emeritus Sci Advisor PF Gold P: 10,429 Complex Powers You're no doubt familiar with Euler's expression exp(i x) = cos(x) + i sin(x) You're probably also familiar that logarithms can be expressed in any base you'd like, like this: loga x = ( logb x ) / ( logb] a ) For example, if your calculator has only log base 10, and you want to compute log2 16, you could enter log10 16 / log10 2 We can put these facts together to good use. To start with, let's try a simple one: express 2i in the a + bi form. We can express 2i as a power of e by solving this equation: 2i = ex i ln 2 = x We've just used the logarithm rule I described above in "reverse." So we've just changed the problem to expressing exp(i ln 2) in a + bi form. Now we can just apply Euler's identity, and we get exp(i ln 2) = cos(ln 2) + i sin(ln 2). Thus 2i = cos(ln 2) + i sin(ln 2), as we wished to find. Now let's try 21 + i. I'm going to skip all the fanfare and just show the steps. 21+i = ex (1+i) ln 2 = x e(1+i) ln 2 = 21+i eln 2 + i ln 2 eln 2 ei ln 2 2 ei ln 2 2 [ cos(ln 2) + i sin(ln 2) ] 2 cos(ln 2) + 2 i sin(ln 2) Hope this helps. - Warren

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