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If ` (a^(2) + 1)^(2)/(2a-i)=x+iy," then when is the value of " x^(2)+y^(2)`? |
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Answer» Given that, `= (a^(2) + 1)^(2)/(2a-i)=x+iy rArr (a^(2) + 1)^(2)/((2a-i)) x+iy` `rArr = ((a^(2) + 1)^(2)(2a + i))/ ((2a + i) (2a + i))=x+iy` `rArr = ((a^(2) + 1)^(2)(2a + i))/(4a^(2)+1)=x+iy` `rArr = (2a(a^(2) + 1)^(2))/(4a^(2)+1) "and" y=(a^(2) + 1)^(2)/(4a^(2)+1) ` `:. x^(2)+y^(2)= 4a[((a^(2) + 1)^(2))/(4a^(2)+1)]^(2)+[((a^(2) + 1)^(2))/(4a^(2)+1)]^(2) ` `((4a^(2)+1)(a^(2)+1)^(4))/(4a^(2)+1)^(2)=(a^(2)+1)^(4)/(4a^(2)+1)^()` |
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