Complex number on a+ib form?

**komodekork** · January 27th 2013, 09:13 AM

Really not sure if this is the right place to post, but how can i write $(-e)^{i\pi}$ on $a+ib$ form?

**Plato** · January 27th 2013, 10:22 AM

Originally Posted by komodekork

Really not sure if this is the right place to post, but how can i write $(-e)^{i\pi}$ on $a+ib$ form?

This is not an easy problem. Recall that complex exponentiation does follow the usual rules that we use in the real numbers.

If each of $z~\&~w$ is a complex number then $w^z=\exp\(z\text{Log}(w))$

Here $\text{Log}(-e)=\ln|-e|+i\pi=1+i\pi$ . That is the principal value.

So you want $[\exp[(i\pi)(1+i\pi)]=~?$

**abender** · January 27th 2013, 10:39 AM

$(-e)^{i\pi} = (-1\cdot e)^{i\pi} = (-1)^{i\pi} \cdot e^{i\pi} = \left((e^{i\pi})^{i\pi}\right) \cdot e^{i\pi} =$ $e^{i^2\pi^2} \cdot e^{i\pi} = e^{-\pi^2} \cdot e^{i\pi} = -e^{-\pi^2}$

Note that I used this identity: $e^{i\pi}=-1$ .

Does this help?

To to hopefully more directly answer the OP's question, recall that the real numbers are a subset of the complex numbers. So, the answer I detailed above is written in the complex notation $a=-e^{-\pi^2}$ and $b=0$ .

**ILikeSerena** · January 27th 2013, 10:51 AM

Originally Posted by abender

$(-e)^{i\pi} = (-1\cdot e)^{i\pi} = (-1)^{i\pi} \cdot e^{i\pi} = \left((e^{i\pi})^{i\pi}\right) \cdot e^{i\pi} =$ $e^{i^2\pi^2} \cdot e^{i\pi} = e^{-\pi^2} \cdot e^{i\pi} = -e^{-\pi^2}$

Note that I used this identity: $e^{i\pi}=-1$ .

Does this help?

This is the so called principal value.

More accurately you have $-1 = e^{i(\pi + 2k\pi)}$ where k is an integer (slight modification of your identity ;)).

If you substitute this, you'll find:

$(-e)^{i\pi} = - e^{-\pi^2(1+2k)}$

which is multi valued (all real) with the principal value you already found.

**komodekork** · January 27th 2013, 11:01 AM

Thank you all, i was really stuck there for a while.

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