- #1
daniel.e2718
- 10
- 0
I'm completely stumped. So is my high-school calculus teacher, but he hasn't done imaginary powers for forty-five years. Hopefully somebody can explain this...
To clarify, I understand the reasoning between the following equation:
[itex]e^{i x}=cos(x)+i sin(x)[/itex]
Now, I need to put some things on the table
First, do you agree that the following is true:
[itex]a^{b c d}=(a^{b c})^d=(a^{b d})^c=(a^{c d})^b=(a^b)^{c d}=(a^c)^{b d}=(a^d)^{b c}[/itex]
for all values of a, b, c, and d where the original expression is defined?
If not, tell me why...
If yes, let's continue.
I will now define a, b, c, and d.
[itex]a=e,b=\pi,c=i,d=\frac{1}{3}[/itex]
Now, we'll go through some of the above equal expressions
[itex]a^{b c d}=e^{\frac{\pi}{3} i}=cos(\frac{\pi}{3})+i sin(\frac{\pi}{3})=\frac{1}{2}+i\frac{\sqrt{3}}{2}=\frac{1+i\sqrt{3}}{2}[/itex]
[itex](a^{b c})^d=(e^{\pi i})^{\frac{1}{3}}=(cos(\pi)+i sin(\pi))^{\frac{1}{3}}=((-1)+i*(0))^{\frac{1}{3}}=(-1)^{\frac{1}{3}}=-1[/itex]
[itex](a^{b d})^c=(e^{\frac{\pi}{3}})^i=(2.849653908\ldots)^i=\frac{1+i\sqrt{3}}{2}[/itex]
[itex](a^{c d})^b=(e^{\frac{i}{3}})^\pi=(cos(\frac{1}{3})+i sin(\frac{1}{3}))^\pi=(0.944956946\ldots+0.327194697\ldots i)^\pi=\frac{1+i\sqrt{3}}{2}[/itex]
I won't do any of the form (a^x)^(y z), because my problem is already present...
If the first expressions I mentioned are indeed equivalent, then why is the second one that I evaluated negative one?
It is completely confusing.
Oh, and hello to the forum for the first time :P
To clarify, I understand the reasoning between the following equation:
[itex]e^{i x}=cos(x)+i sin(x)[/itex]
Now, I need to put some things on the table
First, do you agree that the following is true:
[itex]a^{b c d}=(a^{b c})^d=(a^{b d})^c=(a^{c d})^b=(a^b)^{c d}=(a^c)^{b d}=(a^d)^{b c}[/itex]
for all values of a, b, c, and d where the original expression is defined?
If not, tell me why...
If yes, let's continue.
I will now define a, b, c, and d.
[itex]a=e,b=\pi,c=i,d=\frac{1}{3}[/itex]
Now, we'll go through some of the above equal expressions
[itex]a^{b c d}=e^{\frac{\pi}{3} i}=cos(\frac{\pi}{3})+i sin(\frac{\pi}{3})=\frac{1}{2}+i\frac{\sqrt{3}}{2}=\frac{1+i\sqrt{3}}{2}[/itex]
[itex](a^{b c})^d=(e^{\pi i})^{\frac{1}{3}}=(cos(\pi)+i sin(\pi))^{\frac{1}{3}}=((-1)+i*(0))^{\frac{1}{3}}=(-1)^{\frac{1}{3}}=-1[/itex]
[itex](a^{b d})^c=(e^{\frac{\pi}{3}})^i=(2.849653908\ldots)^i=\frac{1+i\sqrt{3}}{2}[/itex]
[itex](a^{c d})^b=(e^{\frac{i}{3}})^\pi=(cos(\frac{1}{3})+i sin(\frac{1}{3}))^\pi=(0.944956946\ldots+0.327194697\ldots i)^\pi=\frac{1+i\sqrt{3}}{2}[/itex]
I won't do any of the form (a^x)^(y z), because my problem is already present...
If the first expressions I mentioned are indeed equivalent, then why is the second one that I evaluated negative one?
It is completely confusing.
Oh, and hello to the forum for the first time :P