z-2-i-find-arg-z-

Question Number 85668 by Rio Michael last updated on 23/Mar/20

z = 2 + i

find \arg (z)

Commented by mathmax by abdo last updated on 24/Mar/20

∣ z ∣= \sqrt{2^{2} + 1^{2}} = \sqrt{5} \Rightarrow z = \sqrt{5} {\frac{2}{\sqrt{5}} + \frac{1}{\sqrt{5}} i} = r e^{i θ} \Rightarrow

r = \sqrt{5} a n d c o s θ \frac{2}{\sqrt{5}}, s i n θ = \frac{1}{\sqrt{5}} \Rightarrow t a n θ = \frac{1}{2} \Rightarrow θ = a r c t a n (\frac{1}{2}) = \frac{π}{2} - a r c t a n (2)

a r g (z) \equiv \frac{π}{2} - a r c t a n (2) [2 π]

Commented by Rio Michael last updated on 24/Mar/20

thanks sir