Question Number 68122 by TawaTawa last updated on 05/Sep/19
Commented by TawaTawa last updated on 05/Sep/19
$$\mathrm{Please}\:\mathrm{i}\:\mathrm{don}'\mathrm{t}\:\mathrm{understand}\:\mathrm{the}\:\mathrm{workings}\:\mathrm{here}.\:\:\mathrm{Help}\:\mathrm{me}\:\mathrm{explain}\:\mathrm{please} \\ $$
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](Q68130.png)
$${z}^{\mathrm{2018}\:} +\frac{\mathrm{1}}{\boldsymbol{{z}}^{\mathrm{2016}} }+\boldsymbol{{z}}^{\mathrm{2016}} +\frac{\mathrm{1}}{\boldsymbol{{z}}^{\mathrm{2018}} }=\mathrm{0} \\ $$$$\frac{\boldsymbol{{z}}^{\mathrm{4034}} +\mathrm{1}}{\boldsymbol{{z}}^{\mathrm{2016}} }+\frac{\boldsymbol{{z}}^{\mathrm{4034}} +\mathrm{1}}{\boldsymbol{{z}}^{\mathrm{2018}} }=\mathrm{0} \\ $$$$\Rightarrow\left[{z}^{\mathrm{4034}} +\mathrm{1}\right]\left[\frac{\mathrm{1}}{{z}^{\mathrm{2016}} }+\frac{\mathrm{1}}{{z}^{\mathrm{2018}} }\right]=\mathrm{0} \\ $$$$\Rightarrow{Z}^{\mathrm{4034}} +\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\mid{z}\mid^{\mathrm{4034}} =\mathrm{1}\Rightarrow\mid{z}\mid=\mathrm{1} \\ $$$${or}\:\frac{\mathrm{1}}{{z}^{\mathrm{2016}} }+\frac{\mathrm{1}}{{z}^{\mathrm{2018}} }=\mathrm{0}\Rightarrow{Z}^{\mathrm{2}} +\mathrm{1}=\mathrm{0}\Rightarrow\mid{z}\mid=\mathrm{1} \\ $$$$ \\ $$
Commented by TawaTawa last updated on 06/Sep/19
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$