When does the equation (z1z2)^w=(z1^w)(z2^w) hold for all complex values of w?

[Lucy Yeats]

Homework Statement

Verify that the equation (z1z2)^w=(z1^w)(z2^w) is violated for z1=z2=-1 and w=-i.
Under what conditions on the complex values z1 and z2 does the equation hold for all
complex values of w?

Homework Equations

The Attempt at a Solution

((-1)x(-1))^(-i)=1^(-i)=e^ln(1^(-i))=e^(-iln1)=e^0=1

(-1)^(-i)x(-1)^(-i)=(-1)^(-2i)=((-1)^2)^(-i)=1^(-i) but this is the same as above.

Thanks in advance for helping. :-)

 

[I like Serena]

Hi Lucy Yeats!

I'm afraid you can't just use the regular rules for "ln" on complex numbers.
"ln" is not a well defined function for complex numbers.
It's a so called multi-valued function.

You have a similar problem with a^(bc).
It is not just equal to (a^b)^c.
It's more complex. ;)Try using the length-angle representation of a complex number.
That is, $z=r e^{i \phi}$, where $0 \le \phi \lt 2\pi$ and $r \ge 0$.
So try writing -1 as $e^{i \pi}$ and 1 as $e^0$.

 

[Lucy Yeats]

I know that ln(z1z2)=lnz1+lnz2+2πin, where n is an integer.

I don't see why I need to use logs in this question. why can't I say:

(-1x-1)^(-i)=1^(-i)
(-1)^(-i)x(-1)^(-i)=((-1)^(-i))^2=(-1)^(-2i)=((-1)^2)^(-i)=1^(-i)

Thanks for helping!

 

[Lucy Yeats]

Sorry, I see what you mean about the a^bc thing. Ignore the last post and I'll try again.
:-)

 

[I like Serena]

The step
(-1)^(-2i)=((-1)^2)^(-i)
does not hold.

Try the same step with e^(i pi), which is -1.

And no, you should not use logs in this question - at least not on complex numbers, but at most only on real numbers.

Sorry, I see what you mean about the a^bc thing. Ignore the last post and I'll try again.
:-)

EDIT: Too late. ;-)