Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 139057 by EnterUsername last updated on 21/Apr/21

The area of the region in the complex plane satisfying  the inequality log_(cos((π/6))) [((∣z−2∣+5)/(4∣z−2∣−4))]<2 is ?

$$\mathrm{The}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{region}\:\mathrm{in}\:\mathrm{the}\:\mathrm{complex}\:\mathrm{plane}\:\mathrm{satisfying} \\ $$ $$\mathrm{the}\:\mathrm{inequality}\:\mathrm{log}_{\mathrm{cos}\left(\frac{\pi}{\mathrm{6}}\right)} \left[\frac{\mid\mathrm{z}−\mathrm{2}\mid+\mathrm{5}}{\mathrm{4}\mid\mathrm{z}−\mathrm{2}\mid−\mathrm{4}}\right]<\mathrm{2}\:\mathrm{is}\:? \\ $$

Answered by MJS_new last updated on 22/Apr/21

∣z−2∣=x≥0  ((ln ((x+5)/(4(x−1))))/(ln cos (π/6)))<2 ⇔ ln ((x+5)/(x−1)) >ln 3  ((x+5)/(x−1))>3 ⇒ x<4  0≤x<4  0≤∣z−2∣<4  this is a circle with center  ((2),(0) ) and radius 4  but without the circle line

$$\mid{z}−\mathrm{2}\mid={x}\geqslant\mathrm{0} \\ $$ $$\frac{\mathrm{ln}\:\frac{{x}+\mathrm{5}}{\mathrm{4}\left({x}−\mathrm{1}\right)}}{\mathrm{ln}\:\mathrm{cos}\:\frac{\pi}{\mathrm{6}}}<\mathrm{2}\:\Leftrightarrow\:\mathrm{ln}\:\frac{{x}+\mathrm{5}}{{x}−\mathrm{1}}\:>\mathrm{ln}\:\mathrm{3} \\ $$ $$\frac{{x}+\mathrm{5}}{{x}−\mathrm{1}}>\mathrm{3}\:\Rightarrow\:{x}<\mathrm{4} \\ $$ $$\mathrm{0}\leqslant{x}<\mathrm{4} \\ $$ $$\mathrm{0}\leqslant\mid{z}−\mathrm{2}\mid<\mathrm{4} \\ $$ $$\mathrm{this}\:\mathrm{is}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{center}\:\begin{pmatrix}{\mathrm{2}}\\{\mathrm{0}}\end{pmatrix}\:\mathrm{and}\:\mathrm{radius}\:\mathrm{4} \\ $$ $$\mathrm{but}\:\mathrm{without}\:\mathrm{the}\:\mathrm{circle}\:\mathrm{line} \\ $$

Commented byEnterUsername last updated on 22/Apr/21

Thank you Sir

$${Thank}\:{you}\:{Sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com