Question-36385

Question Number 36385 by solihin last updated on 01/Jun/18

Commented by solihin last updated on 01/Jun/18

$$\mathrm{10}.{b}? \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 01/Jun/18

z_1 =3−3i z_2 =3+2i z_1 ^− =3+3i z_1 ^− z_2 =(3+3i)(3+2i) =9+6i+9i+6i^2 =9+15i−6 =3+15i ((z_1 ^− z_2 )/(13))=(3/(13))+((15)/(13))i (i^3 /(−z_1 ))=((−i)/(−3+3i)) =(i/(3−3i))=((i(3+3i))/(3^2 +3^2 ))=((3i−3)/(18)) =((−1)/6)+(i/6) so conjugate of ((−1)/6)+(i/(6 )) is((−1)/6)−(i/6) ((3/(13))+((15)/(13))i)+(((−1)/6)−(i/6)) =((3/(13))−(1/6))+i(((15)/(13))−(1/6)) =(5/(78))+i(((77)/(78))) =

$${z}_{\mathrm{1}} =\mathrm{3}−\mathrm{3}{i}\:\:{z}_{\mathrm{2}} =\mathrm{3}+\mathrm{2}{i} \\ $$$$\overset{−} {{z}}_{\mathrm{1}} =\mathrm{3}+\mathrm{3}{i} \\ $$$$\overset{−} {{z}}_{\mathrm{1}} {z}_{\mathrm{2}} =\left(\mathrm{3}+\mathrm{3}{i}\right)\left(\mathrm{3}+\mathrm{2}{i}\right) \\ $$$$=\mathrm{9}+\mathrm{6}{i}+\mathrm{9}{i}+\mathrm{6}{i}^{\mathrm{2}} \\ $$$$=\mathrm{9}+\mathrm{15}{i}−\mathrm{6} \\ $$$$=\mathrm{3}+\mathrm{15}{i} \\ $$$$\frac{\overset{−} {{z}}_{\mathrm{1}} {z}_{\mathrm{2}} }{\mathrm{13}}=\frac{\mathrm{3}}{\mathrm{13}}+\frac{\mathrm{15}}{\mathrm{13}}{i} \\ $$$$\frac{{i}^{\mathrm{3}} }{−{z}_{\mathrm{1}} }=\frac{−{i}}{−\mathrm{3}+\mathrm{3}{i}} \\ $$$$=\frac{{i}}{\mathrm{3}−\mathrm{3}{i}}=\frac{{i}\left(\mathrm{3}+\mathrm{3}{i}\right)}{\mathrm{3}^{\mathrm{2}} +\mathrm{3}^{\mathrm{2}} }=\frac{\mathrm{3}{i}−\mathrm{3}}{\mathrm{18}} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{6}}+\frac{{i}}{\mathrm{6}} \\ $$$${so}\:{conjugate}\:{of}\:\frac{−\mathrm{1}}{\mathrm{6}}+\frac{{i}}{\mathrm{6}\:}\:\:{is}\frac{−\mathrm{1}}{\mathrm{6}}−\frac{{i}}{\mathrm{6}} \\ $$$$\left(\frac{\mathrm{3}}{\mathrm{13}}+\frac{\mathrm{15}}{\mathrm{13}}{i}\right)+\left(\frac{−\mathrm{1}}{\mathrm{6}}−\frac{{i}}{\mathrm{6}}\right) \\ $$$$=\left(\frac{\mathrm{3}}{\mathrm{13}}−\frac{\mathrm{1}}{\mathrm{6}}\right)+{i}\left(\frac{\mathrm{15}}{\mathrm{13}}−\frac{\mathrm{1}}{\mathrm{6}}\right) \\ $$$$=\frac{\mathrm{5}}{\mathrm{78}}+{i}\left(\frac{\mathrm{77}}{\mathrm{78}}\right) \\ $$$$= \\ $$

Commented by solihin last updated on 02/Jun/18

=(i^3 /(−z_2 ))=((−i)/(−3−2i))=((−i(−3+2i))/(3^2 +2^2 ))=((2+3i)/(13)) =so conjugate of ((2+3i)/(13)) is ((2−3i)/(13))@(2/(13))−(3/(13))i =((3/(13))+((15)/(13))i)+((2/(13))−(3/(13))i) =(5/(13))+((12)/(13))i

$$=\frac{{i}^{\mathrm{3}} }{−{z}_{\mathrm{2}} }=\frac{−{i}}{−\mathrm{3}−\mathrm{2}{i}}=\frac{−{i}\left(−\mathrm{3}+\mathrm{2}{i}\right)}{\mathrm{3}^{\mathrm{2}} +\mathrm{2}^{\mathrm{2}} }=\frac{\mathrm{2}+\mathrm{3}{i}}{\mathrm{13}} \\ $$$$={so}\:{conjugate}\:{of}\:\frac{\mathrm{2}+\mathrm{3}{i}}{\mathrm{13}}\:{is}\:\frac{\mathrm{2}−\mathrm{3}{i}}{\mathrm{13}}@\frac{\mathrm{2}}{\mathrm{13}}−\frac{\mathrm{3}}{\mathrm{13}}{i} \\ $$$$=\left(\frac{\mathrm{3}}{\mathrm{13}}+\frac{\mathrm{15}}{\mathrm{13}}{i}\right)+\left(\frac{\mathrm{2}}{\mathrm{13}}−\frac{\mathrm{3}}{\mathrm{13}}{i}\right) \\ $$$$=\frac{\mathrm{5}}{\mathrm{13}}+\frac{\mathrm{12}}{\mathrm{13}}{i} \\ $$

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