|Jul1-03, 01:20 PM||#1|
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?
|Jul1-03, 01:42 PM||#2|
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.
|Jul1-03, 01:49 PM||#3|
|Jul1-03, 01:51 PM||#4|
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.
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