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Question Number 132937 by mohammad17 last updated on 17/Feb/21
solve the equation in complex number  z^3 =z_o
$${solve}\:{the}\:{equation}\:{in}\:{complex}\:{number} \\ $$$${z}^{\mathrm{3}} ={z}_{{o}} \\ $$
Answered by mr W last updated on 17/Feb/21
z_0 =re^(θi) =re^((2kπ+θ)i)  with k∈Z  ⇒z=(r)^(1/3) e^((((2kπ)/3)+(θ/3))i)  with k=0,1,2
$${z}_{\mathrm{0}} ={re}^{\theta{i}} ={re}^{\left(\mathrm{2}{k}\pi+\theta\right){i}} \:{with}\:{k}\in\mathbb{Z} \\ $$$$\Rightarrow{z}=\sqrt[{\mathrm{3}}]{{r}}{e}^{\left(\frac{\mathrm{2}{k}\pi}{\mathrm{3}}+\frac{\theta}{\mathrm{3}}\right){i}} \:{with}\:{k}=\mathrm{0},\mathrm{1},\mathrm{2} \\ $$
Commented by mohammad17 last updated on 17/Feb/21
sir can you give me steps
$${sir}\:{can}\:{you}\:{give}\:{me}\:{steps} \\ $$
Commented by mr W last updated on 17/Feb/21
there is no more step. it′s a very basic  thing for complex numbers.    with what do you have difficulty?  (1) with why z_0 =re^(θi)   (2) with why re^(θi) =re^((2kπ+θ)i)   (3) with why z=(r)^(1/3) e^((((2kπ)/3)+(θ/3))i)   (4) with why k=0,1,2  and not k∈Z?  (5) all of above  (6) none of above
$${there}\:{is}\:{no}\:{more}\:{step}.\:{it}'{s}\:{a}\:{very}\:{basic} \\ $$$${thing}\:{for}\:{complex}\:{numbers}. \\ $$$$ \\ $$$${with}\:{what}\:{do}\:{you}\:{have}\:{difficulty}? \\ $$$$\left(\mathrm{1}\right)\:{with}\:{why}\:{z}_{\mathrm{0}} ={re}^{\theta{i}} \\ $$$$\left(\mathrm{2}\right)\:{with}\:{why}\:{re}^{\theta{i}} ={re}^{\left(\mathrm{2}{k}\pi+\theta\right){i}} \\ $$$$\left(\mathrm{3}\right)\:{with}\:{why}\:{z}=\sqrt[{\mathrm{3}}]{{r}}{e}^{\left(\frac{\mathrm{2}{k}\pi}{\mathrm{3}}+\frac{\theta}{\mathrm{3}}\right){i}} \\ $$$$\left(\mathrm{4}\right)\:{with}\:{why}\:{k}=\mathrm{0},\mathrm{1},\mathrm{2}\:\:{and}\:{not}\:{k}\in\mathbb{Z}? \\ $$$$\left(\mathrm{5}\right)\:{all}\:{of}\:{above} \\ $$$$\left(\mathrm{6}\right)\:{none}\:{of}\:{above} \\ $$

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