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Question Number 146235 by mathdanisur last updated on 12/Jul/21

if   arg z_1  = ((4π)/3)   and   arg z_1 ^2  ∙ z = ((7π)/6)  find   arg z = ?

$${if}\:\:\:{arg}\:{z}_{\mathrm{1}} \:=\:\frac{\mathrm{4}\pi}{\mathrm{3}}\:\:\:{and}\:\:\:{arg}\:{z}_{\mathrm{1}} ^{\mathrm{2}} \:\centerdot\:{z}\:=\:\frac{\mathrm{7}\pi}{\mathrm{6}} \\ $$$${find}\:\:\:{arg}\:{z}\:=\:? \\ $$

Answered by gsk2684 last updated on 12/Jul/21

arg z_1 ^2 z=2arg z_1 +arg z+2kΠ

$${arg}\:{z}_{\mathrm{1}} ^{\mathrm{2}} {z}=\mathrm{2}{arg}\:{z}_{\mathrm{1}} +{arg}\:{z}+\mathrm{2}{k}\Pi \\ $$

Answered by puissant last updated on 12/Jul/21

arg z_1 ^2 .z = arg z_1 ^2 +arg z = ((7π)/6)  ⇒arg z=((7π)/6)−2arg z_1 =((7π)/6)−((8π)/3)  ⇒ arg z=−((3π)/2)..

$$\mathrm{arg}\:\mathrm{z}_{\mathrm{1}} ^{\mathrm{2}} .\mathrm{z}\:=\:\mathrm{arg}\:\mathrm{z}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{arg}\:\mathrm{z}\:=\:\frac{\mathrm{7}\pi}{\mathrm{6}} \\ $$$$\Rightarrow\mathrm{arg}\:\mathrm{z}=\frac{\mathrm{7}\pi}{\mathrm{6}}−\mathrm{2arg}\:\mathrm{z}_{\mathrm{1}} =\frac{\mathrm{7}\pi}{\mathrm{6}}−\frac{\mathrm{8}\pi}{\mathrm{3}} \\ $$$$\Rightarrow\:\mathrm{arg}\:\mathrm{z}=−\frac{\mathrm{3}\pi}{\mathrm{2}}.. \\ $$

Commented by mathdanisur last updated on 12/Jul/21

cool Ser, thanks

$${cool}\:{Ser},\:{thanks}\: \\ $$

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