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if-arg-z-i-i-2-find-Imz-Rez-




Question Number 146318 by mathdanisur last updated on 12/Jul/21
if   arg (((z−i)/i))=2  find   Imz + Rez = ?
$${if}\:\:\:{arg}\:\left(\frac{{z}−{i}}{{i}}\right)=\mathrm{2} \\ $$$${find}\:\:\:{Imz}\:+\:{Rez}\:=\:? \\ $$
Commented by Ar Brandon last updated on 12/Jul/21
−π≤arg(Z)≤π
$$−\pi\leqslant\mathrm{arg}\left(\mathrm{Z}\right)\leqslant\pi \\ $$
Commented by mathdanisur last updated on 12/Jul/21
this is answer Ser.?
$${this}\:{is}\:{answer}\:{Ser}.? \\ $$
Commented by Ar Brandon last updated on 12/Jul/21
No ! I just don′t understand  your “2” since −π≤arg(Z)≤π
$$\mathrm{No}\:!\:\mathrm{I}\:\mathrm{just}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand} \\ $$$$\mathrm{your}\:“\mathrm{2}''\:\mathrm{since}\:−\pi\leqslant\mathrm{arg}\left(\mathrm{Z}\right)\leqslant\pi \\ $$
Commented by mr W last updated on 12/Jul/21
no unique value for Imz+Rez!
$${no}\:{unique}\:{value}\:{for}\:{Imz}+{Rez}! \\ $$
Answered by mr W last updated on 12/Jul/21
let z=x+yi  ((z−i)/i)=((zi−i^2 )/i^2 )=−1−zi=−1−(x+yi)i  =y−1−xi  arg(((z−i)/i))=arg(y−1−xi)=2  y−1<0, x<0  ((−x)/(1−y))=tan (π−2)  ⇒y=(x/(tan (π−2)))+1 with x<0  the locus of z is the line y=(x/(tan (π−2)))+1  with x<0.    Imz+Rez=y+x≠constant
$${let}\:{z}={x}+{yi} \\ $$$$\frac{{z}−{i}}{{i}}=\frac{{zi}−{i}^{\mathrm{2}} }{{i}^{\mathrm{2}} }=−\mathrm{1}−{zi}=−\mathrm{1}−\left({x}+{yi}\right){i} \\ $$$$={y}−\mathrm{1}−{xi} \\ $$$${arg}\left(\frac{{z}−{i}}{{i}}\right)={arg}\left({y}−\mathrm{1}−{xi}\right)=\mathrm{2} \\ $$$${y}−\mathrm{1}<\mathrm{0},\:{x}<\mathrm{0} \\ $$$$\frac{−{x}}{\mathrm{1}−{y}}=\mathrm{tan}\:\left(\pi−\mathrm{2}\right) \\ $$$$\Rightarrow{y}=\frac{{x}}{\mathrm{tan}\:\left(\pi−\mathrm{2}\right)}+\mathrm{1}\:{with}\:{x}<\mathrm{0} \\ $$$${the}\:{locus}\:{of}\:{z}\:{is}\:{the}\:{line}\:{y}=\frac{{x}}{\mathrm{tan}\:\left(\pi−\mathrm{2}\right)}+\mathrm{1} \\ $$$${with}\:{x}<\mathrm{0}. \\ $$$$ \\ $$$${Imz}+{Rez}={y}+{x}\neq{constant} \\ $$
Commented by mathdanisur last updated on 13/Jul/21
Thank you Ser, answer: y+x.?
$${Thank}\:{you}\:{Ser},\:{answer}:\:{y}+{x}.? \\ $$
Commented by mr W last updated on 13/Jul/21
i have said: value of x+y is not  unique!
$${i}\:{have}\:{said}:\:{value}\:{of}\:{x}+{y}\:{is}\:{not} \\ $$$${unique}! \\ $$

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