Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 196265 by KRIMO last updated on 21/Aug/23

Answered by a.lgnaoui last updated on 21/Aug/23

posons z=a+ib ⇔z=(√(a^2 +b^2  )) ((a/( (√(a^2 +b^2 ))))+i(b/( (√(a^2 +b^2 )))))  avec  (a/( (√(a^2 +b^2 ))))=cos α     (b/( (√(a^2 +b^2 ))))=sin α  z=(1+(√3) )(1+i)  ici  a=b=1  (√(a^2 +b^2 )) =(√2)  z=(√2)(1+(√3) )((1/( (√2)))+i(1/( (√2))))=((√2)+(√6) ) (cos (π/4)+isin (( π)/4))

$$\mathrm{posons}\:\mathrm{z}=\mathrm{a}+\mathrm{ib}\:\Leftrightarrow\mathrm{z}=\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} \:}\:\left(\frac{\mathrm{a}}{\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }}+\mathrm{i}\frac{\mathrm{b}}{\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }}\right) \\ $$$$\mathrm{avec}\:\:\frac{\mathrm{a}}{\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }}=\mathrm{cos}\:\alpha\:\:\:\:\:\frac{\mathrm{b}}{\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }}=\mathrm{sin}\:\alpha \\ $$$$\mathrm{z}=\left(\mathrm{1}+\sqrt{\mathrm{3}}\:\right)\left(\mathrm{1}+\mathrm{i}\right) \\ $$$$\mathrm{ici}\:\:\mathrm{a}=\mathrm{b}=\mathrm{1}\:\:\sqrt{\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} }\:=\sqrt{\mathrm{2}} \\ $$$$\mathrm{z}=\sqrt{\mathrm{2}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\:\right)\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{i}\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)=\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{6}}\:\right)\:\left(\mathrm{cos}\:\frac{\pi}{\mathrm{4}}+\mathrm{isin}\:\frac{\:\pi}{\mathrm{4}}\right) \\ $$$$ \\ $$

Answered by Rodier97 last updated on 22/Aug/23

   posons Z=(1+(√3))+i(1+(√3))    Alors ∣Z∣=(√((1+(√3))^2 +(1+(√3))^2 ))=(√(2(1+(√3))^2 ))               ∣Z∣=(√2)(1+(√3))  Et:   { ((cosθ=((1+(√3))/( (√2)(1+(√(3))))) =((√2)/2))),((sinθ=((1+(√3) )/( (√2)(1+(√3))))=((√2)/2))) :} ⇒  θ=(π/4)    D′ou^�    Z=(√2)(1+(√3))[cos(π/4)+i sin(π/4)]

$$ \\ $$$$\:\mathrm{posons}\:\mathrm{Z}=\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)+\boldsymbol{\mathrm{i}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right) \\ $$$$ \\ $$$$\mathrm{Alors}\:\mid\mathrm{Z}\mid=\sqrt{\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} +\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }=\sqrt{\mathrm{2}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} } \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\mid\mathrm{Z}\mid=\sqrt{\mathrm{2}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right) \\ $$$$\mathrm{Et}: \\ $$$$\begin{cases}{\mathrm{cos}\theta=\frac{\mathrm{1}+\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{2}}\left(\mathrm{1}+\sqrt{\left.\mathrm{3}\right)}\right.}\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\\{\mathrm{sin}\theta=\frac{\mathrm{1}+\sqrt{\mathrm{3}}\:}{\:\sqrt{\mathrm{2}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}\end{cases}\:\Rightarrow\:\:\theta=\frac{\pi}{\mathrm{4}} \\ $$$$ \\ $$$$\mathrm{D}'\mathrm{o}\grave {\mathrm{u}}\:\:\:\mathrm{Z}=\sqrt{\mathrm{2}}\left(\mathrm{1}+\sqrt{\mathrm{3}}\right)\left[\mathrm{cos}\frac{\pi}{\mathrm{4}}+\boldsymbol{\mathrm{i}}\:\mathrm{sin}\frac{\pi}{\mathrm{4}}\right] \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com