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Question Number 69794 by mathmax by abdo last updated on 27/Sep/19

$$letp\left(x\right)={(x+1)}^6-e^{i\alpha }with\alpha real$$$$1)findtherootsofp\left(x\right)$$$$2)factorizep\left(x\right)insideC\left[x\right]$$$$3)factorizep\left(x\right)insideR\left[x\right]$$

Commented by mathmax by abdo last updated on 01/Oct/19

$$1)p\left(x\right)=0\iff {(x+1)}^6=e^{i\alpha }\iff \frac{{(x+1)}^6}{e^{i\alpha }}=1\iff \frac{{(x+1)}^6}{{\left(e^{\frac{i\alpha }{6}}\right)}^6}=1\Rightarrow$$$${\left((x+1)e^{-\frac{i\alpha }{6}}\right)}^6=e^{i2k\pi }\Rightarrow (x+1)e^{-\frac{i\alpha }{6}}=e^{\frac{i2k\pi }{6}}\Rightarrow$$$$(x+1)=e^{\frac{ik\pi }{3}}{e}^{\frac{i\alpha }{6}}\Rightarrow x+1=e^{\frac{i(2k\pi +\alpha )}{6}}\Rightarrow x=e^{i(\frac{2k\pi +\alpha )}{6}}-1$$$$sotherootsofp\left(x\right)are{z}_k=e^{i\left(\frac{2k\pi +\alpha }{6}\right)}-1withk\in \left[[0,5]\right]$$$$2)p\left(x\right)=a\prod _{k=0}^5(x-z_k)weseethata=1\Rightarrow$$$$p\left(x\right)=\prod _{k=0}^5(x+1-e^{i\left(\frac{3k\pi +\alpha }{6}\right)})$$

Commented by mathmax by abdo last updated on 01/Oct/19

$$erroroftypop\left(x\right)=\prod _{k=0}^5(x+1-e^{i\left(\frac{2k\pi +\alpha }{6}\right)})$$

Answered by mind is power last updated on 29/Sep/19

$$p\left(x\right)=0\Rightarrow$$$${(x+1)}^6=e^{ia}\iff {(x+1)}^6=e^{i(a+2k\pi )}$$$$\Rightarrow x+1=e^{i(a+2k\pi )/6},k\in [0,5]$$$$\Rightarrow x_k=-1+e^{i\frac{a+2k\pi }{6}}$$$$2)p\left(x\right)=\underset{k=0}{\prod ^5}(x-(-1+e^{i\frac{a+2k\pi }{6}}))$$$$p\left(x\right)=\underset{k=0}{\prod ^5}(x+1-e^{i\frac{a+2k\pi }{6}})$$$$overIRnotpossibleonly[if{e}^{ia}\in IR\Rightarrow a=k\pi$$$$causep(-1)=-e^{ia}$$$$$$$$$$

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