- #1

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^{1+i}) in the form a+bi, or some other standard form of representation for complex numbers?

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- Thread starter Lonewolf
- Start date

- #1

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- #2

HallsofIvy

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Specfically, look at e

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

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2^{1+i}= 2*2i

Now why didn't I see that? Oh well, thanks for pointing it out.

- #4

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:

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

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

For example, if your calculator has only log base 10, and you want to compute log

log

We can put these facts together to good use.

2

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

2

(1+i) ln 2 = x

e

e

e

2 e

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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