Complex Powers

  • Thread starter Lonewolf
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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?
 

HallsofIvy

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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.
 
334
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21+i= 2*2i
Now why didn't I see that? Oh well, thanks for pointing it out. :smile:
 

chroot

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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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