Please check my calculations with these complex numbers

[MatinSAR]

Homework Statement

Calculate the following terms.

Relevant Equations

Complex numbers.

$$Re(e^{2iz}) = Re(\cos(2z)+i\sin(2z))=\cos(2z)$$$$e^{i^3} = e^{-i}$$ $$\ln (\sqrt 3 + i)^3=\ln(2)+i(\dfrac {\pi}{6}+2k\pi)$$
Can't I simplify these more? Are they correct?

Final one:$(1+3i)^{\frac 1 2}$
Can I write in in term of $\sin x$ and $\cos x$ then use $(\cos x+i\sin x)^n=\cos (nx)+i\sin (nx)$?
Something like this :

Many thanks.

 

[Hill]

Check this one: $\ln (\sqrt 3 + i)^3=\ln(2)+i(\dfrac {\pi}{6}+2k\pi)$. I suspect you forgot about the power 3.

 

[MatinSAR]

Check this one: $\ln (\sqrt 3 + i)^3=\ln(2)+i(\dfrac {\pi}{6}+2k\pi)$. I suspect you forgot about the power 3.

Yes. Thanks.
$$\ln (\sqrt 3 + i)^3=\ln (2e^{i (\frac {\pi}{6}+2k\pi) })^3=\ln (8e^{3i (\frac {\pi}{6}+2k\pi) })=\ln(8)+3i (\frac {\pi}{6}+2k\pi) $$

 

[FactChecker]

Homework Statement: Calculate the following terms.
Relevant Equations: Complex numbers.

$$Re(e^{2iz}) = Re(\cos(2z)+i\sin(2z))=\cos(2z)$$

It looks like you are assuming that $z$ is a real number. Otherwise, $\cos(2z)$ is not necessarily real.
If it is real, you should state that (and I would prefer that you use $r$ instead of $z$).

 

[MatinSAR]

@Hill @FactChecker
Thanks for your time.

 

[FactChecker]

IMO, some of these would be simpler if they were transformed using Euler's equation: $e^{i r} = \cos(r)+i\sin(r)$ for any real number, $r$.

$$e^{i^3} = e^{-i}$$

Wouldn't that be easier to visualize as $\cos(-1)+i\sin(-1) = \cos(1)-i\sin(1)$?

Final one:$(1+3i)^{\frac 1 2}$
Can I write in in term of $\sin x$ and $\cos x$ then use $(\cos x+i\sin x)^n=\cos (nx)+i\sin (nx)$?

What if you change it to polar coordinates using Euler's formula? Powers are much easier in that form.
$(r e^{ix})^{\frac 1 2} = r^{\frac 1 2} e^{i{\frac 1 2}x}$.
Then you can use Euler's formula again to put the result back to Cartesian coordinates if you want.

 

[MatinSAR]

Wouldn't that be easier to visualize as $\cos(-1)+i\sin(-1) = \cos(1)-i\sin(1)$?

How $\cos(-1)+i\sin(-1) = \cos(1)-i\sin(1)$ is possible? I did not understand.

What if you change it to polar coordinates using Euler's formula? Powers are much easier in that form.
$(r e^{ix})^{\frac 1 2} = r^{\frac 1 2} e^{i{\frac 1 2}x}$.
Then you can use Euler's formula again to put the result back to Cartesian coordinates if you want.

Thanks for the suggestion. All my answers were correct except the one you've mentioend in post #4. That $z$ is a complex number so my answer is wrong.

 

[FactChecker]

How $\cos(-1)+i\sin(-1) = \cos(1)-i\sin(1)$ is possible? I did not understand.

cos(x) is an even function, so cos(-x) = cos(x).
sin(x) is an odd function, so sin(-x) = -sin(x).

Thanks for the suggestion. All my answers were correct except the one you've mentioend in post #4. That $z$ is a complex number so my answer is wrong.

So express z as $a+ib, a,b\in R$.
Then $e^{2iz} = e^{2i(a+ib)} = e^{i(2a)} e^{-2b}$.
See what you can do from there.

 

[MatinSAR]

cos(x) is an even function, so cos(-x) = cos(x).
sin(x) is an odd function, so sin(-x) = -sin(x).

Idk how I didn't remember this. Perhaps because it's too late here.

cos(x) is an even function, so cos(-x) = cos(x).
sin(x) is an odd function, so sin(-x) = -sin(x).

So express z as $a+ib, a,b\in R$.
Then $e^{2iz} = e^{2i(a+ib)} = e^{i(2a)} e^{-2b}$.
See what you can do from there.

I will try tomorrow. Thanks for your time.

 

[MatinSAR]

So express z as $a+ib, a,b\in R$.
Then $e^{2iz} = e^{2i(a+ib)} = e^{i(2a)} e^{-2b}$.
See what you can do from there.

$$e^{2iz} = e^{2i(a+ib)} = e^{i(2a)} e^{-2b}$$$$Re(e^{2iz}) = e^{-2b} cos(2a)$$

 

[FactChecker]

$$e^{2iz} = e^{2i(a+ib)} = e^{i(2a)} e^{-2b}$$$$Re(e^{2iz}) = e^{-2b} cos(2a)$$

Good. I might have gone farther in my hint than I should have. Some people say that Euler's Formula is the most important formula in mathematics. In any case, it is important to get used to using it to go back and forth between polar and Cartesian coordinates. It can make things a lot easier.

 

[MatinSAR]

Good. I might have gone farther in my hint than I should have. Some people say that Euler's Formula is the most important formula in mathematics. In any case, it is important to get used to using it to go back and forth between polar and Cartesian coordinates. It can make things a lot easier.

Thanks for your time @FactChecker .