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Question Number 149142 by mathdanisur last updated on 03/Aug/21

if z + āˆ£zāˆ£ = 1 + (āˆš3) i find š›Ÿ = ?

$${if}\:\:\:{z}\:+\:\mid{z}\mid\:=\:\mathrm{1}\:+\:\sqrt{\mathrm{3}}\:\boldsymbol{{i}} \\ $$$${find}\:\:\:\boldsymbol{\varphi}\:=\:? \\ $$

Answered by liberty last updated on 03/Aug/21

z=a+bi āˆ£zāˆ£=(āˆš(a^2 +b^2 )) ā‡’a+bi+(āˆš(a^2 +b^2 )) = 1+(āˆš3) i { ((b=(āˆš3))),((a+(āˆš(a^2 +3)) =1 ā‡’(āˆš(a^2 +3)) =1āˆ’a)) :} ā‡’a^2 +3=a^2 āˆ’2a+1 ā‡’a=āˆ’1 then z=āˆ’1+(āˆš3) i tan Ļ•=((āˆš3)/(āˆ’1))=āˆ’(āˆš3) Ļ•=āˆ’(Ļ€/3)

$$\mathrm{z}=\mathrm{a}+\mathrm{b}{i} \\ $$$$\mid\mathrm{z}\mid=\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} } \\ $$$$\Rightarrow\mathrm{a}+\mathrm{b}{i}+\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }\:=\:\mathrm{1}+\sqrt{\mathrm{3}}\:{i} \\ $$$$\begin{cases}{{b}=\sqrt{\mathrm{3}}}\\{{a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{3}}\:=\mathrm{1}\:\Rightarrow\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{3}}\:=\mathrm{1}āˆ’\mathrm{a}}\end{cases} \\ $$$$\Rightarrow\mathrm{a}^{\mathrm{2}} +\mathrm{3}=\mathrm{a}^{\mathrm{2}} āˆ’\mathrm{2a}+\mathrm{1} \\ $$$$\Rightarrow\mathrm{a}=āˆ’\mathrm{1}\:\mathrm{then}\:\mathrm{z}=āˆ’\mathrm{1}+\sqrt{\mathrm{3}}\:{i} \\ $$$$\mathrm{tan}\:\varphi=\frac{\sqrt{\mathrm{3}}}{āˆ’\mathrm{1}}=āˆ’\sqrt{\mathrm{3}}\: \\ $$$$\varphi=āˆ’\frac{\pi}{\mathrm{3}} \\ $$

Commented by mathdanisur last updated on 03/Aug/21

Cool, Thank You Ser

$${Cool},\:{Thank}\:{You}\:{Ser} \\ $$

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