Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 68122 by TawaTawa last updated on 05/Sep/19

Commented by TawaTawa last updated on 05/Sep/19

Please i don′t understand the workings here.  Help me explain please

$$\mathrm{Please}\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:\mathrm{the}\:\mathrm{workings}\:\mathrm{here}.\:\:\mathrm{Help}\:\mathrm{me}\:\mathrm{explain}\:\mathrm{please} \\ $$

Answered by mind is power last updated on 05/Sep/19

z^(2018 ) +(1/z^(2016) )+z^(2016) +(1/z^(2018) )=0  ((z^(4034) +1)/z^(2016) )+((z^(4034) +1)/z^(2018) )=0  ⇒[z^(4034) +1][(1/z^(2016) )+(1/z^(2018) )]=0  ⇒Z^(4034) +1=0  ⇒∣z∣^(4034) =1⇒∣z∣=1  or (1/z^(2016) )+(1/z^(2018) )=0⇒Z^2 +1=0⇒∣z∣=1

$${z}^{\mathrm{2018}\:} +\frac{\mathrm{1}}{\boldsymbol{{z}}^{\mathrm{2016}} }+\boldsymbol{{z}}^{\mathrm{2016}} +\frac{\mathrm{1}}{\boldsymbol{{z}}^{\mathrm{2018}} }=\mathrm{0} \\ $$$$\frac{\boldsymbol{{z}}^{\mathrm{4034}} +\mathrm{1}}{\boldsymbol{{z}}^{\mathrm{2016}} }+\frac{\boldsymbol{{z}}^{\mathrm{4034}} +\mathrm{1}}{\boldsymbol{{z}}^{\mathrm{2018}} }=\mathrm{0} \\ $$$$\Rightarrow\left[{z}^{\mathrm{4034}} +\mathrm{1}\right]\left[\frac{\mathrm{1}}{{z}^{\mathrm{2016}} }+\frac{\mathrm{1}}{{z}^{\mathrm{2018}} }\right]=\mathrm{0} \\ $$$$\Rightarrow{Z}^{\mathrm{4034}} +\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\mid{z}\mid^{\mathrm{4034}} =\mathrm{1}\Rightarrow\mid{z}\mid=\mathrm{1} \\ $$$${or}\:\frac{\mathrm{1}}{{z}^{\mathrm{2016}} }+\frac{\mathrm{1}}{{z}^{\mathrm{2018}} }=\mathrm{0}\Rightarrow{Z}^{\mathrm{2}} +\mathrm{1}=\mathrm{0}\Rightarrow\mid{z}\mid=\mathrm{1} \\ $$$$ \\ $$

Commented by TawaTawa last updated on 06/Sep/19

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com