Problematic Proof?

Algebra Level 5

Any complex number $z$ , with $|z|=1$ can be expressed in the form $e^{i\theta +2ki\pi}$ .
$(a^{b})^{c} = a^{bc}$ meaning that $z^{i}=e^{i*(i\theta +2ki\pi)} = e^{-\theta -2k\pi}$

Result: $z^{i}$ holds an infinite number of values all with argument 0 but with a range of magnitudes.

Is there a problem with this and if so with which step does it lie?

The proof is correct Result is inferred incorrectly Step 1 Step 2

1 solution

William Whitehouse
Feb 23, 2017

The proof is correct: in the same way that $4^{0.5}$ is both 2 and -2 any complex number magnitude 1 raised to a complex power can be an infinite number of different things

4^0.5 is only 2. -4^0.5 is -2.

J D - 4 years, 3 months ago

Just to clarify, with -4^0.5 I meant -(4^0.5)

J D - 4 years, 3 months ago

No, 4^0.5 is both things (and can be proven to be both as above) the square root of 4 is, as you say, only 2 as the square root function always returns the principle root (i.e k=0)

William Whitehouse - 4 years, 3 months ago

As a real exponential $4^{0.5} = 2$ , not $-2$ . This is dictated by enforcing continuity on exponential functions.

As a complex exponential, you are correct that $4^{0.5}$ must be allowed to be both $2, -2$ until we choose a branch.

These two exponentials are very different creatures and we should make sure to not confuse them...

Brian Moehring - 4 years, 3 months ago

Problematic Proof?

1 solution

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