A) \[\frac{-2{{b}^{3}}}{{{({{a}^{2}}+{{b}^{2}})}^{2}}}\]
B) \[\frac{6{{a}^{2}}b}{{{({{a}^{2}}+{{b}^{2}})}^{2}}}\]
C) 0
D) \[\frac{5{{a}^{2}}b}{{{a}^{2}}+{{b}^{2}}}\]
Correct Answer: C
Solution :
[c] If \[\frac{{{\left( a+ib \right)}^{2}}}{a-ib~~}-\frac{{{\left( a-ib \right)}^{2}}}{a+ib}=x+iy\] \[\Rightarrow \frac{{{\left( a+ib \right)}^{3}}{{\left( a-ib \right)}^{3}}}{\left( a-ib~ \right)\left( a+ib \right)~}=x+iy\] \[\left[ \because \text{ }{{a}^{3}}-{{b}^{3}}={{\left( a-b \right)}^{3}}+3ab\left( a-b \right) \right]\] \[{{\left( a+ib-a+ib \right)}^{3}}+3.\left( a+ib \right)\] \[\Rightarrow \frac{(a-ib)(a+ib-a+ib)}{{{a}^{2}}+{{b}^{2}}}=x+iy\] \[\Rightarrow \frac{-8ib+3\left( {{a}^{2}}+{{b}^{2}} \right)\left( 2bi \right)}{{{a}^{2}}+{{b}^{2}}}=x+iy\] L.H.S. of the above result be purely imginary. Hence, Rel. part \[x=0\] Hence, option [c] is correct.
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