# Please check my calculations with these complex numbers

• MatinSAR
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.

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

Hill
MatinSAR said:
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
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##.
MatinSAR said:
$$e^{i^3} = e^{-i}$$
Wouldn't that be easier to visualize as ##\cos(-1)+i\sin(-1) = \cos(1)-i\sin(1)##?
MatinSAR said:
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
FactChecker said:
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.
FactChecker said:
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.

MatinSAR said:
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).
MatinSAR said:
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
FactChecker said:
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.
FactChecker said:
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.

FactChecker said:
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
MatinSAR said:
$$e^{2iz} = e^{2i(a+ib)} = e^{i(2a)} e^{-2b}$$$$Re(e^{2iz}) = e^{-2b} cos(2a)$$
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
FactChecker said:
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 .

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