How do we express complex powers like 21+i in standard form?

First, of course, 2¹⁺ⁱ= 2*2ⁱ so the question is really about 2ⁱ (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.

2¹⁺ⁱ= 2*2i

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

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 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 = 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 2ⁱ = 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.

2¹⁺ⁱ = e^x
(1+i) ln 2 = x

e^{(1+i) ln 2} = 2¹⁺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