What-is-the-argument-of-the-complex-numbers-below-i-z-1-e-pi-6-i-ii-z-1-e-pi-6-i-

Question Number 145524 by physicstutes last updated on 05/Jul/21

What is the argument of the complex numbers below

(i) z = 1 + e^{\frac{π}{6} i}

(ii) z = 1 - e^{\frac{π}{6} i}

Answered by Olaf_Thorendsen last updated on 05/Jul/21

(i) z = 1 + e^{i \frac{π}{6}}

z = e^{i \frac{π}{12}} (e^{- i \frac{π}{12}} + e^{i \frac{π}{12}})

z = 2 \cos \frac{π}{12} e^{i \frac{π}{12}}

\arg z = \frac{π}{12} [2 π]

(ii) z = 1 - e^{\frac{π}{6} i}

z = e^{i \frac{π}{12}} (e^{- i \frac{π}{12}} - e^{i \frac{π}{12}})

z = - 2 i \sin \frac{π}{12} e^{i \frac{π}{12}}

z = e^{i π} 2 e^{i \frac{π}{2}} \sin \frac{π}{12} e^{i \frac{π}{12}}

z = 2 \sin \frac{π}{12} e^{i (π + \frac{π}{2} + \frac{π}{12})}

z = 2 \sin \frac{π}{12} e^{i \frac{19 π}{12}}

\arg z = \frac{19 π}{12} [2 π]

Commented by physicstutes last updated on 05/Jul/21

Thanks sir .