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Question Number 80890 by M±th+et£s last updated on 07/Feb/20

$${(\frac{1+\mathrm{i}}{2-\mathrm{i}}+\frac{2+\mathrm{i}}{1-\mathrm{i}})}^3=500x+500y\mathrm{i}$$$$findx,y$$

Answered by behi83417@gmail.com last updated on 07/Feb/20

$$\frac{(1+\mathrm{i})(2+\mathrm{i})}{2+1}=\frac{2+\mathrm{i}+2\mathrm{i}+1}{3}=\frac{3+3\mathrm{i}}{3}=1+\mathrm{i}$$$$\frac{2+\mathrm{i}}{1-\mathrm{i}}=\frac{(2+\mathrm{i})(1+\mathrm{i})}{1+1}=\frac{2+2\mathrm{i}+\mathrm{i}+1}{2}=\frac{3+3\mathrm{i}}{2}$$$$\mathrm{LHS}={[1+\mathrm{i}+\frac{3}{2}(1+\mathrm{i})]}^3={\left[\frac{5}{2}(1+\mathrm{i})\right]}^3=$$$$=\frac{125}{8}(1+3\mathrm{i}+{3\mathrm{i}}^2+{\mathrm{i}}^3)=\frac{125}{8}(1+3\mathrm{i}-3-\mathrm{i})=$$$$=\frac{125}{8}(2\mathrm{i}-2)=\frac{125}{4}(\mathrm{i}-1)=500\mathrm{x}+500\mathrm{yi}$$$$\Rightarrow \mathrm{i}-1=16\mathrm{x}+16\mathrm{yi}\Rightarrow \{\begin{array}{l}\mathrm{x}=\frac{-1}{16}\\ \mathrm{y}=\frac{1}{16}\end{array}.$$

Commented by M±th+et£s last updated on 07/Feb/20

$$thankyousir$$

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