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Question Number 33463 by NECx last updated on 17/Apr/18

Find the pricipal and ordinary  argument of z=(i/(−2−2i))

$${Find}\:{the}\:{pricipal}\:{and}\:{ordinary} \\ $$$${argument}\:{of}\:{z}=\frac{{i}}{−\mathrm{2}−\mathrm{2}{i}} \\ $$

Commented by abdo imad last updated on 17/Apr/18

z = ((−i)/(2(1+i))) = ((−i(1−i))/4) = ((1−i)/4) ⇒∣z∣ =(1/4)∣1−i∣ =((√2)/4)  z =(1/4) −(i/4) =((√2)/4)( (1/(√2)) −(i/(√2))) =((√2)/2) e^(−i(π/4))  ⇒arg(z)≡−(π/4)[2π] .

$${z}\:=\:\frac{−{i}}{\mathrm{2}\left(\mathrm{1}+{i}\right)}\:=\:\frac{−{i}\left(\mathrm{1}−{i}\right)}{\mathrm{4}}\:=\:\frac{\mathrm{1}−{i}}{\mathrm{4}}\:\Rightarrow\mid{z}\mid\:=\frac{\mathrm{1}}{\mathrm{4}}\mid\mathrm{1}−{i}\mid\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}} \\ $$$${z}\:=\frac{\mathrm{1}}{\mathrm{4}}\:−\frac{{i}}{\mathrm{4}}\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}\left(\:\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:−\frac{{i}}{\sqrt{\mathrm{2}}}\right)\:=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:{e}^{−{i}\frac{\pi}{\mathrm{4}}} \:\Rightarrow{arg}\left({z}\right)\equiv−\frac{\pi}{\mathrm{4}}\left[\mathrm{2}\pi\right]\:. \\ $$

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