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Question Number 132632 by mohammad17 last updated on 15/Feb/21

Solve the equation z^3 =z_° z=x+iy ,z_° =(x_° +iy_° )

$${Solve}\:{the}\:{equation}\:{z}^{\mathrm{3}} ={z}_{°} \\ $$$${z}={x}+{iy}\:,{z}_{°} =\left({x}_{°} +{iy}_{°} \right) \\ $$

Answered by TheSupreme last updated on 15/Feb/21

x_0 +iy_0 =ρe^(iθ) ρ=(√(x_0 ^2 +y_0 ^2 )) tan(θ)=(y_0 /x_0 ) we have to correct phase θ=atan((x_0 /y_0 ))+π sgn(x_0 ) z=z_0 ^(1/3) =ρ^(1/3) e^(i ((θ+2nπ)/3)) z_n =(x_0 ^2 +y_0 ^2 )^(1/6) e^(i((atan((x_0 /y_0 ))+π(2n+sgn(x_0 )))/3)) i=0,1,2

$${x}_{\mathrm{0}} +{iy}_{\mathrm{0}} =\rho{e}^{{i}\theta} \:\rho=\sqrt{{x}_{\mathrm{0}} ^{\mathrm{2}} +{y}_{\mathrm{0}} ^{\mathrm{2}} }\:{tan}\left(\theta\right)=\frac{{y}_{\mathrm{0}} }{{x}_{\mathrm{0}} } \\ $$$${we}\:{have}\:{to}\:{correct}\:{phase} \\ $$$$\theta={atan}\left(\frac{{x}_{\mathrm{0}} }{{y}_{\mathrm{0}} }\right)+\pi\:{sgn}\left({x}_{\mathrm{0}} \right) \\ $$$${z}={z}_{\mathrm{0}} ^{\frac{\mathrm{1}}{\mathrm{3}}} =\rho^{\frac{\mathrm{1}}{\mathrm{3}}} {e}^{{i}\:\frac{\theta+\mathrm{2}{n}\pi}{\mathrm{3}}} \\ $$$${z}_{{n}} =\left({x}_{\mathrm{0}} ^{\mathrm{2}} +{y}_{\mathrm{0}} ^{\mathrm{2}} \right)^{\frac{\mathrm{1}}{\mathrm{6}}} {e}^{{i}\frac{{atan}\left(\frac{{x}_{\mathrm{0}} }{{y}_{\mathrm{0}} }\right)+\pi\left(\mathrm{2}{n}+{sgn}\left({x}_{\mathrm{0}} \right)\right)}{\mathrm{3}}} \:{i}=\mathrm{0},\mathrm{1},\mathrm{2} \\ $$$$ \\ $$

Commented by mohammad17 last updated on 15/Feb/21

sir can you give me defintion of sgn please

$${sir}\:{can}\:{you}\:{give}\:{me}\:{defintion}\:{of}\:{sgn}\:{please} \\ $$

Commented by Ar Brandon last updated on 15/Feb/21

sgn(x)=sign of x

$$\mathrm{sgn}\left(\mathrm{x}\right)=\mathrm{sign}\:\mathrm{of}\:\mathrm{x} \\ $$

Commented by mr W last updated on 15/Feb/21

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