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Complex Powers 
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#1
Jul103, 01:20 PM

P: 333

How do we express complex powers of numbers (e.g. 2^{1+i}) in the form a+bi, or some other standard form of representation for complex numbers?



#2
Jul103, 01:42 PM

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PF Gold
P: 39,353

First, of course, 2^{1+i}= 2*2^{i} so the question is really about 2^{i} (or, more generally, a^{bi}).
Specfically, look at e^{ix}. 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. 


#3
Jul103, 01:49 PM

P: 333




#4
Jul103, 01:51 PM

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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: log_{a} x = ( log_{b} x ) / ( log_{b]} a ) For example, if your calculator has only log base 10, and you want to compute log_{2} 16, you could enter log_{10} 16 / log_{10} 2 We can put these facts together to good use. To start with, let's try a simple one: express 2^{i} in the a + bi form. We can express 2^{i} as a power of e by solving this equation: 2^{i} = e^{x} 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 2^{i} = cos(ln 2) + i sin(ln 2), as we wished to find. Now let's try 2^{1 + i}. I'm going to skip all the fanfare and just show the steps. 2^{1+i} = e^{x} (1+i) ln 2 = x e^{(1+i) ln 2} = 2^{1+i} e^{ln 2 + i ln 2} e^{ln 2} e^{i ln 2} 2 e^{i 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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