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Question Number 68515 by azizullah last updated on 12/Sep/19

Commented by Prithwish sen last updated on 13/Sep/19

[z(z−6)]^2 =[(3+2i)(3+2i−6)]^2 =[(2i)^2 −(3)^2 ]^2 =(−13)^2 =169

$$\left[\mathrm{z}\left(\mathrm{z}−\mathrm{6}\right)\right]^{\mathrm{2}} =\left[\left(\mathrm{3}+\mathrm{2i}\right)\left(\mathrm{3}+\mathrm{2i}−\mathrm{6}\right)\right]^{\mathrm{2}} =\left[\left(\mathrm{2i}\right)^{\mathrm{2}} −\left(\mathrm{3}\right)^{\mathrm{2}} \right]^{\mathrm{2}} \\ $$$$=\left(−\mathrm{13}\right)^{\mathrm{2}} =\mathrm{169} \\ $$

Commented by azizullah last updated on 13/Sep/19

thanks

$$\mathrm{thanks} \\ $$

Answered by Rasheed.Sindhi last updated on 12/Sep/19

=z^2 (z^2 −12z+36) =(3+2i)^2 ((3+2i)^2 −12(3+2i)+36) =(9−4+12i)(9−4+12i−36−24i+36) =(5+12i)(5+12i−24i) =(5+12i)(5−12i) =(5)^2 −(12i)^2 =25+144=169

$$=\mathrm{z}^{\mathrm{2}} \left(\mathrm{z}^{\mathrm{2}} −\mathrm{12z}+\mathrm{36}\right) \\ $$$$=\left(\mathrm{3}+\mathrm{2i}\right)^{\mathrm{2}} \left(\left(\mathrm{3}+\mathrm{2i}\right)^{\mathrm{2}} −\mathrm{12}\left(\mathrm{3}+\mathrm{2i}\right)+\mathrm{36}\right) \\ $$$$=\left(\mathrm{9}−\mathrm{4}+\mathrm{12i}\right)\left(\mathrm{9}−\mathrm{4}+\mathrm{12i}−\mathrm{36}−\mathrm{24i}+\mathrm{36}\right) \\ $$$$=\left(\mathrm{5}+\mathrm{12i}\right)\left(\mathrm{5}+\mathrm{12i}−\mathrm{24i}\right) \\ $$$$=\left(\mathrm{5}+\mathrm{12i}\right)\left(\mathrm{5}−\mathrm{12i}\right) \\ $$$$=\left(\mathrm{5}\right)^{\mathrm{2}} −\left(\mathrm{12i}\right)^{\mathrm{2}} \\ $$$$=\mathrm{25}+\mathrm{144}=\mathrm{169} \\ $$

Commented by azizullah last updated on 13/Sep/19

sir! thanks of good work.

$$\boldsymbol{\mathrm{sir}}!\:\boldsymbol{\mathrm{thanks}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{good}}\:\boldsymbol{\mathrm{work}}. \\ $$

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