What is the argument of the complex numbers below (i) z = 1+e^((π/6)i) (ii) z = 1 −e^((π/6)i)


Question and Answers Forum

All Questions Topic List
Logarithms Questions
Previous in All Question Next in All Question
Previous in Logarithms Next in Logarithms
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

	$Thanks sir .$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum