Menu Close

z-z-1-3-i-find-




Question Number 207563 by hardmath last updated on 18/May/24
z  +  ∣z∣  =  1  +  (√3) i  find:   𝛟 = ?
$$\mathrm{z}\:\:+\:\:\mid\mathrm{z}\mid\:\:=\:\:\mathrm{1}\:\:+\:\:\sqrt{\mathrm{3}}\:\boldsymbol{\mathrm{i}} \\ $$$$\mathrm{find}:\:\:\:\boldsymbol{\varphi}\:=\:? \\ $$
Answered by Frix last updated on 18/May/24
z+∣z∣=1+(√3)i  ∣z∣=1+(√3)i−z  ∣z∣∈R ⇒ z=a+(√3)i  ∣a+(√3)i∣=1−a  (√(a^2 +3))=1−a ⇒ a=−1  z=−1+(√3)i=2e^(i((2π)/3))  ⇒ ϕ=((2π)/3)
$${z}+\mid{z}\mid=\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i} \\ $$$$\mid{z}\mid=\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i}−{z} \\ $$$$\mid{z}\mid\in\mathbb{R}\:\Rightarrow\:{z}={a}+\sqrt{\mathrm{3}}\mathrm{i} \\ $$$$\mid{a}+\sqrt{\mathrm{3}}\mathrm{i}\mid=\mathrm{1}−{a} \\ $$$$\sqrt{{a}^{\mathrm{2}} +\mathrm{3}}=\mathrm{1}−{a}\:\Rightarrow\:{a}=−\mathrm{1} \\ $$$${z}=−\mathrm{1}+\sqrt{\mathrm{3}}\mathrm{i}=\mathrm{2e}^{\mathrm{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \:\Rightarrow\:\varphi=\frac{\mathrm{2}\pi}{\mathrm{3}} \\ $$
Commented by hardmath last updated on 19/May/24
thank you professor
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{professor} \\ $$
Answered by A5T last updated on 18/May/24
z=a+bi⇒∣z∣=(√(a^2 +b^2 ))  z+∣z∣=(a+(√(a^2 +b^2 )))+bi=1+i(√3)⇒b=(√3)  a+(√(a^2 +b^2 ))=1⇒a+(√(a^2 +3))=1⇒a^2 +3=1+a^2 −2a  ⇒a=−1⇒z=−1+i(√3)  ⇒z=2e^(i((2π)/3))
$${z}={a}+{bi}\Rightarrow\mid{z}\mid=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${z}+\mid{z}\mid=\left({a}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\right)+{bi}=\mathrm{1}+{i}\sqrt{\mathrm{3}}\Rightarrow{b}=\sqrt{\mathrm{3}} \\ $$$${a}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }=\mathrm{1}\Rightarrow{a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{3}}=\mathrm{1}\Rightarrow{a}^{\mathrm{2}} +\mathrm{3}=\mathrm{1}+{a}^{\mathrm{2}} −\mathrm{2}{a} \\ $$$$\Rightarrow{a}=−\mathrm{1}\Rightarrow{z}=−\mathrm{1}+{i}\sqrt{\mathrm{3}} \\ $$$$\Rightarrow{z}=\mathrm{2}{e}^{{i}\frac{\mathrm{2}\pi}{\mathrm{3}}} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *