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

$$\left(1\right)\mathrm{find}\mathrm{the}\mathrm{locus}\mid \mathrm{z}-{\mathrm{z}}_1\mid =2\mathrm{meets}$$$$\mathrm{the}\mathrm{positive}\mathrm{real}\mathrm{axis}$$$$\left(2\right)\mathrm{On}\mathrm{a}\mathrm{single}\mathrm{Argand}\mathrm{diagram},\mathrm{sketch}$$$$\mathrm{the}\mathrm{loci}\to \{\begin{array}{l}\mid \mathrm{z}-{\mathrm{z}}_1\mid =2\\ \arg (\mathrm{z}-{\mathrm{z}}_2)=\frac{\pi }{4}\end{array}$$

Commented by MJS_new last updated on 08/Sep/20

$$\left(2\right)\mathrm{we}\mathrm{cannot}\mathrm{scetch}\mathrm{the}\mathrm{loci}\mathrm{if}\mathrm{we}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{know}$$$$\mathrm{any}\mathrm{coordinates}.$$$$\mid z-z_1\mid =2\mathrm{is}\mathrm{the}\mathrm{equations}\mathrm{of}\mathrm{all}\mathrm{circles}\mathrm{with}$$$$\mathrm{radius}2$$$$\arg (z-z_2)=\frac{\pi }{4}\mathrm{is}\mathrm{any}\mathrm{straight}\mathrm{line}\mathrm{with}\mathrm{slope}45°$$$$\mathrm{let}z-z_2=r{\mathrm{e}}^{\mathrm{i}\theta }\Rightarrow \arg r{\mathrm{e}}^{\mathrm{i}\theta }=\theta \Rightarrow \theta =\frac{\pi }{4}$$$$\arg ((x-p)+(y-q)\mathrm{i})=\frac{\pi }{4}$$$$\mathrm{leads}\mathrm{to}y=x-p+q$$

Answered by MJS_new last updated on 08/Sep/20

$$\left(1\right)$$$$\mathrm{assuming}{z}_1\mathrm{is}\mathrm{given}\mathrm{and}z\mathrm{is}\mathrm{variable}\mathrm{we}\mathrm{have}$$$$z_1=p+q\mathrm{i}\wedge z=x+y\mathrm{i}$$$$\mid z-z_1\mid =2$$$$\mid (x-p)+(y-q)\mathrm{i}\mid =2$$$$\sqrt{{(x-p)}^2+{(y-q)}^2}=2$$$$\mathrm{both}\mathrm{sides}\mathrm{are}⩾0\Rightarrow \mathrm{we}\mathrm{are}\mathrm{allowed}\mathrm{to}\mathrm{square}$$$${(x-p)}^2+{(y-q)}^2=4$$$$\mathrm{which}\mathrm{is}\mathrm{a}\mathrm{circle}\mathrm{with}\mathrm{center}\left(\begin{array}{c}p\\ q\end{array}\right)\mathrm{and}\mathrm{radius}4$$$$\mathrm{if}\mathrm{it}\mathrm{meets}\mathrm{the}\mathrm{positive}\mathrm{real}\mathrm{axis}\mathrm{depends}\mathrm{on}$$$$\mathrm{the}\mathrm{values}\mathrm{if}p\mathrm{and}q$$$$-2⩽q⩽2\Rightarrow \mathrm{it}\mathrm{meets}\mathrm{the}\mathrm{real}\mathrm{axis}$$$$p>2\Rightarrow \mathrm{it}\mathrm{meets}\mathrm{the}\mathrm{positive}\mathrm{real}\mathrm{axis}$$$$\mathrm{we}\mathrm{find}\mathrm{the}\mathrm{intersection}\mathrm{point}\left(\mathrm{s}\right)\mathrm{putting}$$$$y=0$$$${(x-p)}^2+q^2=4$$$$\Rightarrow x=p\pm \sqrt{4-q^2}\mathrm{with}p>2\wedge -2⩽q⩽2$$

Commented by bemath last updated on 08/Sep/20

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

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