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Question Number 146318 by mathdanisur last updated on 12/Jul/21

$$ifarg\left(\frac{z-i}{i}\right)=2$$$$findImz+Rez=?$$

Commented by Ar Brandon last updated on 12/Jul/21

$$-\pi ⩽\arg \left(\mathrm{Z}\right)⩽\pi$$

Commented by mathdanisur last updated on 12/Jul/21

$$thisisanswerSer.?$$

Commented by Ar Brandon last updated on 12/Jul/21

$$\mathrm{No}!\mathrm{I}\mathrm{just}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{understand}$$$$\mathrm{your}‘‘2^{″}\mathrm{since}-\pi ⩽\arg \left(\mathrm{Z}\right)⩽\pi$$

Commented by mr W last updated on 12/Jul/21

$$nouniquevalueforImz+Rez!$$

Answered by mr W last updated on 12/Jul/21

$$letz=x+yi$$$$\frac{z-i}{i}=\frac{zi-i^2}{i^2}=-1-zi=-1-(x+yi)i$$$$=y-1-xi$$$$arg\left(\frac{z-i}{i}\right)=arg(y-1-xi)=2$$$$y-1<0,x<0$$$$\frac{-x}{1-y}=\tan (\pi -2)$$$$\Rightarrow y=\frac{x}{\tan (\pi -2)}+1withx<0$$$$thelocusofzistheliney=\frac{x}{\tan (\pi -2)}+1$$$$withx<0.$$$$$$$$Imz+Rez=y+x\ne constant$$

Commented by mathdanisur last updated on 13/Jul/21

$$ThankyouSer,answer:y+x.?$$

Commented by mr W last updated on 13/Jul/21

$$ihavesaid:valueofx+yisnot$$$$unique!$$

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