Question and Answers Forum

All Questions      Topic List

Others Questions

Previous in All Question      Next in All Question      

Previous in Others      Next in Others      

Question Number 112454 by bemath last updated on 08/Sep/20

(1) find the locus ∣z−z_1 ∣ = 2 meets the positive real axis (2)On a single Argand diagram, sketch the loci → { ((∣z−z_1 ∣=2)),((arg(z−z_2 )=(π/4))) :}

$$\left(\mathrm{1}\right)\:\mathrm{find}\:\mathrm{the}\:\mathrm{locus}\:\mid\mathrm{z}−\mathrm{z}_{\mathrm{1}} \mid\:=\:\mathrm{2}\:\mathrm{meets} \\ $$$$\mathrm{the}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{axis} \\ $$$$\left(\mathrm{2}\right)\mathrm{On}\:\mathrm{a}\:\mathrm{single}\:\mathrm{Argand}\:\mathrm{diagram},\:\mathrm{sketch} \\ $$$$\mathrm{the}\:\mathrm{loci}\:\rightarrow\begin{cases}{\mid\mathrm{z}−\mathrm{z}_{\mathrm{1}} \mid=\mathrm{2}}\\{\mathrm{arg}\left(\mathrm{z}−\mathrm{z}_{\mathrm{2}} \right)=\frac{\pi}{\mathrm{4}}}\end{cases} \\ $$

Commented by MJS_new last updated on 08/Sep/20

(2) we cannot scetch the loci if we don′t know any coordinates. ∣z−z_1 ∣=2 is the equations of all circles with radius 2 arg (z−z_2 ) =(π/4) is any straight line with slope 45° let z−z_2 =re^(iθ) ⇒ arg re^(iθ) =θ ⇒ θ=(π/4) arg ((x−p)+(y−q)i) =(π/4) leads to y=x−p+q

$$\left(\mathrm{2}\right)\:\mathrm{we}\:\mathrm{cannot}\:\mathrm{scetch}\:\mathrm{the}\:\mathrm{loci}\:\mathrm{if}\:\mathrm{we}\:\mathrm{don}'\mathrm{t}\:\mathrm{know} \\ $$$$\mathrm{any}\:\mathrm{coordinates}. \\ $$$$\mid{z}−{z}_{\mathrm{1}} \mid=\mathrm{2}\:\mathrm{is}\:\mathrm{the}\:\mathrm{equations}\:\mathrm{of}\:\mathrm{all}\:\mathrm{circles}\:\mathrm{with} \\ $$$$\mathrm{radius}\:\mathrm{2} \\ $$$$\mathrm{arg}\:\left({z}−{z}_{\mathrm{2}} \right)\:=\frac{\pi}{\mathrm{4}}\:\mathrm{is}\:\mathrm{any}\:\mathrm{straight}\:\mathrm{line}\:\mathrm{with}\:\mathrm{slope}\:\mathrm{45}° \\ $$$$\mathrm{let}\:{z}−{z}_{\mathrm{2}} ={r}\mathrm{e}^{\mathrm{i}\theta} \:\Rightarrow\:\mathrm{arg}\:{r}\mathrm{e}^{\mathrm{i}\theta} \:=\theta\:\Rightarrow\:\theta=\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{arg}\:\left(\left({x}−{p}\right)+\left({y}−{q}\right)\mathrm{i}\right)\:=\frac{\pi}{\mathrm{4}} \\ $$$$\mathrm{leads}\:\mathrm{to}\:{y}={x}−{p}+{q} \\ $$

Answered by MJS_new last updated on 08/Sep/20

(1) assuming z_1 is given and z is variable we have z_1 =p+qi ∧ z=x+yi ∣z−z_1 ∣=2 ∣(x−p)+(y−q)i∣=2 (√((x−p)^2 +(y−q)^2 ))=2 both sides are ≥0 ⇒ we are allowed to square (x−p)^2 +(y−q)^2 =4 which is a circle with center ((p),(q) ) and radius 4 if it meets the positive real axis depends on the values if p and q −2≤q≤2 ⇒ it meets the real axis p>2 ⇒ it meets the positive real axis we find the intersection point(s) putting y=0 (x−p)^2 +q^2 =4 ⇒ x=p±(√(4−q^2 )) with p>2∧−2≤q≤2

$$\left(\mathrm{1}\right) \\ $$$$\mathrm{assuming}\:{z}_{\mathrm{1}} \:\mathrm{is}\:\mathrm{given}\:\mathrm{and}\:{z}\:\mathrm{is}\:\mathrm{variable}\:\mathrm{we}\:\mathrm{have} \\ $$$${z}_{\mathrm{1}} ={p}+{q}\mathrm{i}\:\wedge\:{z}={x}+{y}\mathrm{i} \\ $$$$\mid{z}−{z}_{\mathrm{1}} \mid=\mathrm{2} \\ $$$$\mid\left({x}−{p}\right)+\left({y}−{q}\right)\mathrm{i}\mid=\mathrm{2} \\ $$$$\sqrt{\left({x}−{p}\right)^{\mathrm{2}} +\left({y}−{q}\right)^{\mathrm{2}} }=\mathrm{2} \\ $$$$\mathrm{both}\:\mathrm{sides}\:\mathrm{are}\:\geqslant\mathrm{0}\:\Rightarrow\:\mathrm{we}\:\mathrm{are}\:\mathrm{allowed}\:\mathrm{to}\:\mathrm{square} \\ $$$$\left({x}−{p}\right)^{\mathrm{2}} +\left({y}−{q}\right)^{\mathrm{2}} =\mathrm{4} \\ $$$$\mathrm{which}\:\mathrm{is}\:\mathrm{a}\:\mathrm{circle}\:\mathrm{with}\:\mathrm{center}\:\begin{pmatrix}{{p}}\\{{q}}\end{pmatrix}\:\:\mathrm{and}\:\mathrm{radius}\:\mathrm{4} \\ $$$$\mathrm{if}\:\mathrm{it}\:\mathrm{meets}\:\mathrm{the}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{axis}\:\mathrm{depends}\:\mathrm{on} \\ $$$$\mathrm{the}\:\mathrm{values}\:\mathrm{if}\:{p}\:\mathrm{and}\:{q} \\ $$$$−\mathrm{2}\leqslant{q}\leqslant\mathrm{2}\:\Rightarrow\:\mathrm{it}\:\mathrm{meets}\:\mathrm{the}\:\mathrm{real}\:\mathrm{axis} \\ $$$${p}>\mathrm{2}\:\Rightarrow\:\mathrm{it}\:\mathrm{meets}\:\mathrm{the}\:\mathrm{positive}\:\mathrm{real}\:\mathrm{axis} \\ $$$$\mathrm{we}\:\mathrm{find}\:\mathrm{the}\:\mathrm{intersection}\:\mathrm{point}\left(\mathrm{s}\right)\:\mathrm{putting} \\ $$$${y}=\mathrm{0} \\ $$$$\left({x}−{p}\right)^{\mathrm{2}} +{q}^{\mathrm{2}} =\mathrm{4} \\ $$$$\Rightarrow\:{x}={p}\pm\sqrt{\mathrm{4}−{q}^{\mathrm{2}} }\:\mathrm{with}\:{p}>\mathrm{2}\wedge−\mathrm{2}\leqslant{q}\leqslant\mathrm{2} \\ $$

Commented by bemath last updated on 08/Sep/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com