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Question Number 32002 by rahul 19 last updated on 18/Mar/18

If  z^3 =z^�  prove   then ∣z∣=1.

$${If}\:\:\boldsymbol{{z}}^{\mathrm{3}} =\bar {\boldsymbol{{z}}}\:{prove}\: \\ $$$${then}\:\mid\boldsymbol{{z}}\mid=\mathrm{1}. \\ $$

Answered by Tinkutara last updated on 18/Mar/18

z^3 =z^�   ∣z^3 ∣=∣z^� ∣  ∣z∣^3 =∣z∣  ∣z∣^2 =1  ∣z∣=1

$${z}^{\mathrm{3}} =\bar {{z}} \\ $$$$\mid{z}^{\mathrm{3}} \mid=\mid\bar {{z}}\mid \\ $$$$\mid{z}\mid^{\mathrm{3}} =\mid{z}\mid \\ $$$$\mid{z}\mid^{\mathrm{2}} =\mathrm{1} \\ $$$$\mid{z}\mid=\mathrm{1} \\ $$

Commented by Tinkutara last updated on 18/Mar/18

∣z∣ is always≥0.  If z=x+iy  then ∣z∣=(√(x^2 +y^2 )), where both x,y∈R  and square root of real number is  taken as real number only.

$$\mid{z}\mid\:{is}\:{always}\geqslant\mathrm{0}. \\ $$$${If}\:{z}={x}+{iy} \\ $$$${then}\:\mid{z}\mid=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} },\:{where}\:{both}\:{x},{y}\in{R} \\ $$$${and}\:{square}\:{root}\:{of}\:{real}\:{number}\:{is} \\ $$$${taken}\:{as}\:{real}\:{number}\:{only}. \\ $$

Commented by rahul 19 last updated on 18/Mar/18

thanku!

$${thanku}! \\ $$

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