complex numbers: [e^{3pi i / 2} - e^{5pi i / 6}] / [e^{pi/2} - e^{5pi i / 6}]

[selariathas]

hello! i need help for a calcul :

(e^((i3π)/2) - e^((i5π)/6))/(e^((i3π)/2) - e^((i5π)/6))

[imath]\large{ \dfrac{e^{\frac{3\pi i}{2}} - e^{\frac{5\pi i}{6}}}{e^{\frac{\pi i}{2}} - e^{\frac{5\pi i}{6}}} }[/imath]

 

[stapel]

hello! i need help for a calcul :

[imath]\large{ \dfrac{e^{\frac{3\pi i}{2}} - e^{\frac{5\pi i}{6}}}{e^{\frac{\pi i}{2}} - e^{\frac{5\pi i}{6}}} }[/imath]

Do you mean that you are trying to do a "calculation"? If so, what subjects have you recently studied? (This information can help us try to figure out how you're expected to proceed.) What have you tried so far? (This information will help us see where you're having trouble.)

Thank you!

Eliz.

 

[topsquark]

hello! i need help for a calcul :

(e^((i3π)/2) - e^((i5π)/6))/(e^((i3π)/2) - e^((i5π)/6))

[imath]\large{ \dfrac{e^{\frac{3\pi i}{2}} - e^{\frac{5\pi i}{6}}}{e^{\frac{\pi}{2}} - e^{\frac{5\pi i}{6}}} }[/imath]

Is this statement correct? I'm guessing there should be an i in the exponent of the first term of the denominator.

Once that is there I can probably "simplify" the expression, but I have doubts about actually calculating it. (Admittedly, I have not finished the problem.)

-Dan

 

[khansaheb]

hello! i need help for a calcul :

(e^((i3π)/2) - e^((i5π)/6))/(e^((i3π)/2) - e^((i5π)/6))

[imath]\large{ \dfrac{e^{\frac{3\pi i}{2}} - e^{\frac{5\pi i}{6}}}{e^{\frac{\pi}{2}} - e^{\frac{5\pi i}{6}}} }[/imath]

Have you tried to use Euler's "CiS" formula?

 

[stapel]

Is this statement correct? I'm guessing there should be an i in the exponent of the first term of the denominator.

Typo corrected. Thank you!

Eliz.