InterviewSolution
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दी गयी संख्याओं को `x+iy` के रूप में लिखे । (i) `z=(2+i)/((1+i)(1-2i))` (ii) `((1+i)^(3))/(4+3i)` |
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Answer» (i) `z=(2+i)/((1+i)*(1-2i))=(2+i)/(3-i)=((2+i)(3+i))/((3-i)(3+i))=(5+5i)/(10)=(1)/(2)+(i)/(2)` (ii) `z=((1+i)^(3))/(4+3i)=(1^(3)+3*1^(2)*i+3+i^(2)+i^(3))/(4+3i)=(1+3i-3-i)/(4+3i)` `=(-2+2i)/(4+3i)=((-2+2i)(4-3i))/((4+3i)(4-3i))` `=(-8+8i+6i-6i^(2))/(4^(2)-9i^(2))=(-2+14i)/(25)=-(2)/(25)+i(14)/(25)` |
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