Complex Analysis Properties Question 2

[RJLiberator]

The problem states, Show that:
a) |e^(i*theta)| = 1.
Now, the definition of e^(i*theta) makes this
|cos(theta)+isin(theta)|
If we choose any theta then this should be equal to 1.

What can help me prove this? If I choose, say, pi/6 then it simplifies to |(sqrt(3))/2+i/2)| which doesn't seem to equal 1.

b) BAR(e^(i*theta)) = e^(-i*theta)

Here's what I think. The bar e^(i*theta) means that the definition is cos(theta)-isin(theta) and this can be rewritten as cos(-theta)+i*sin(-theta) which can be inputted back into the definition to see that e^(-i*theta) is correct.

 

[Dick]

The problem states, Show that:
a) |e^(i*theta)| = 1.
Now, the definition of e^(i*theta) makes this
|cos(theta)+isin(theta)|
If we choose any theta then this should be equal to 1.

What can help me prove this? If I choose, say, pi/6 then it simplifies to |(sqrt(3))/2+i/2)| which doesn't seem to equal 1.

b) BAR(e^(i*theta)) = e^(-i*theta)

Here's what I think. The bar e^(i*theta) means that the definition is cos(theta)-isin(theta) and this can be rewritten as cos(-theta)+i*sin(-theta) which can be inputted back into the definition to see that e^(-i*theta) is correct.

Uh, $|x+iy|=\sqrt{x^2+y^2}$, yes? So?

 

[RJLiberator]

Uh, $|x+iy|=\sqrt{x^2+y^2}$, yes? So?

Ok, so when we have |cos(theta)+isin(theta)| this can be represented as sqrt(cos^2(theta)+sin^2(theta)) and this is clearly equal to 1, always.

Ah, that is simply beautiful.
I appreciate your guidance here. Swiftly helped me here.

 

[micromass]

|(sqrt(3))/2+i/2)| which doesn't seem to equal 1.

It does.

 

[WWGD]

Also, I am not sure what the starting point is, what you can assume, but $cos\theta+ isin\theta$ is a parametrization for a point in the unit circle.