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

Commented by solihin last updated on 01/Jun/18

$$10.b?$$

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

$$z_1=3-3i{z}_2=3+2i$$$${\stackrel{-}{z}}_1=3+3i$$$${\stackrel{-}{z}}_1{z}_2=(3+3i)(3+2i)$$$$=9+6i+9i+6i^2$$$$=9+15i-6$$$$=3+15i$$$$\frac{{\stackrel{-}{z}}_1{z}_2}{13}=\frac{3}{13}+\frac{15}{13}i$$$$\frac{i^3}{-z_1}=\frac{-i}{-3+3i}$$$$=\frac{i}{3-3i}=\frac{i(3+3i)}{3^2+3^2}=\frac{3i-3}{18}$$$$=\frac{-1}{6}+\frac{i}{6}$$$$soconjugateof\frac{-1}{6}+\frac{i}{6}is\frac{-1}{6}-\frac{i}{6}$$$$(\frac{3}{13}+\frac{15}{13}i)+(\frac{-1}{6}-\frac{i}{6})$$$$=(\frac{3}{13}-\frac{1}{6})+i(\frac{15}{13}-\frac{1}{6})$$$$=\frac{5}{78}+i\left(\frac{77}{78}\right)$$$$=$$

Commented by solihin last updated on 02/Jun/18

$$=\frac{i^3}{-z_2}=\frac{-i}{-3-2i}=\frac{-i(-3+2i)}{3^2+2^2}=\frac{2+3i}{13}$$$$=soconjugateof\frac{2+3i}{13}is\frac{2-3i}{13}@\frac{2}{13}-\frac{3}{13}i$$$$=(\frac{3}{13}+\frac{15}{13}i)+(\frac{2}{13}-\frac{3}{13}i)$$$$=\frac{5}{13}+\frac{12}{13}i$$

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