Complex numbers: show that (a^b)^c has more values than a^(bc)

[applestrudle]

Homework Statement

Show that (a^b)^c can have more values than a^(bc)

Use [(-i)^(2+i)]^(2-i) and (-i)^5 or (i^i)^i and i^-1 to show this.

Homework Equations

The Attempt at a Solution

I'm writing out the second one as the first one is long:

i^i = e^ilni

lni = i (\pi /2 ± k2\pi)

so then

i^i = e^(=\pi/2 ±k2\pi)

(i^i)^i = e^i(\pi/2 ± k2\pi)

as e^(±ik2\pi) = 1

you get (i^i)^i = e^i \pi/2

which is one value only (equal to i^-1 which is -i)

this is not what the question suggests



please help!

 

[R136a1]

I don't see why you calculated $(i^i)^i$. The question in the OP asks you to calculate something entirely different.

 

[haruspex]

I don't see why you calculated $(i^i)^i$. The question in the OP asks you to calculate something entirely different.

It does? How?

Applestrudle, I can't see a flaw in what you did. i tried the other pair of expressions and also drew a blank.

 

[applestrudle]

It does? How?

Applestrudle, I can't see a flaw in what you did. i tried the other pair of expressions and also drew a blank.

I thought is was a strange question. Just to confirm, for complex numbers the same laws for real numbers for exponents holds right? so for complex a,b,c (a^bc) = (a^b)^b ?

 

[mfb]

I thought is was a strange question. Just to confirm, for complex numbers the same laws for real numbers for exponents holds right? so for complex a,b,c (a^bc) = (a^b)^b ?

Neglecting the typo, (a^b)^c can have more ambiguity as it can depend more on the choice of the branch for the logarithm.

$$(-i)^{2+i} = exp\left((2+i)(-i \frac{\pi}{2} + 2k \pi i)\right)$$
$$\left((-i)^{2+i}\right)^{2-i} = exp\left((2-i)((2+i)(-i \frac{\pi}{2} + 2k \pi i) +{\color{red}{2l \pi i}})\right) $$
The red part should give additional solutions.

(i^i)^i = e^i(π/2 ± k2π)

Same here:
(i^i)^i = e^i(π/2 ± k2π ± l 2 π i[/color] )

 

[D H]

I thought is was a strange question. Just to confirm, for complex numbers the same laws for real numbers for exponents holds right? so for complex a,b,c (a^bc) = (a^b)^b ?

I assume you meant (a^b)^c = a^bc. That identity is valid only for positive real a and real b and c. The ultimate point of this exercise is to show that this identity no longer holds for complex a, b, and c.

Another one to watch out for is the identity (ab)^c = a^cb^c. This, too, is no longer valid for complex a, b, and c.

 

[applestrudle]

Neglecting the typo, (a^b)^c can have more ambiguity as it can depend more on the choice of the branch for the logarithm.

$$(-i)^{2+i} = exp\left((2+i)(-i \frac{\pi}{2} + 2k \pi i)\right)$$

i swear you can't do that?

 

[mfb]

Sure you can?

 

[applestrudle]

Sure you can?

oh yeah because e is real, I'm new to this sorry.

 

[applestrudle]

what if a and b are real but c is complex?

I assume you meant (a^b)^c = a^bc. That identity is valid only for positive real a and real b and c. The ultimate point of this exercise is to show that this identity no longer holds for complex a, b, and c.

Another one to watch out for is the identity (ab)^c = a^cb^c. This, too, is no longer valid for complex a, b, and c.

For (ab)^c = a^cb^c if a and b are real but c is complex would it still hold?

 

[D H]

For (ab)^c = a^cb^c if a and b are real but c is complex would it still hold?

No. That identity is valid if a and b are non-negative and c is real. All bets are off otherwise.

 

[applestrudle]

No. That identity is valid if a and b are non-negative and c is real. All bets are off otherwise.

are you sure you don't mean a and b are real and c in non-negative?

 

[mfb]

I think "non-negative" implies real, and negatice c are fine if a and b are (real) positive numbers.

 

[D H]

are you sure you don't mean a and b are real and c in non-negative?

No. I meant what I wrote. Try a=b=-1 and c=1/2. It doesn't work. (ab)^1/2 is not equal to a^1/2b^1/2.

 

[SBP]

How does the red part come into play $$2l\pi i$$?
$$((-i)^{(2+i)})^{(2-i)} = e^{((2-i)((2+i)(-i\pi/2 + 2k\pi i) +2l\pi i))}$$

 

[mfb]

You need the logarithm of -i and the complex logarithm has many branches that differ by 2l pi i where l is an integer.

 

[SBP]

Isn't that what the i2kpi is for? Why can you add the second i2lpi when raise it to a complex number?

 

[mfb]

It has the same reason. You have to use the a^b = exp(b ln(a)) conversion twice.

 

[SBP]

I see. Thanks!