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NCERT Class XI Mathematics - Complex Numbers and Quadratic Equations - Solutions

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Question : 14 of 52
Marks: +1, -0
Express the following expression in the form of a + ib :
(3+i5)(3−i5)(3+2i)−(3−i2)\frac{(3+i\sqrt{5})(3-i\sqrt{5})}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})}
Solution:  
We have, (3+i5)(3−i5)(3+2i)−(3−i2)\frac{(3+i\sqrt{5})(3-i\sqrt{5})}{(\sqrt{3}+\sqrt{2}i)-(\sqrt{3}-i\sqrt{2})}
= (3)2−(i5)23+2i−3+2i\frac{(3)^2-(i\sqrt{5})^2}{\sqrt{3}+\sqrt{2}i-\sqrt{3}+\sqrt{2}i} = 9−5i222i\frac{9-5i^2}{2\sqrt{2}i} = 9−5(−1)22i\frac{9-5(-1)}{2\sqrt{2}i} = 9+522i\frac{9+5}{2\sqrt{2}i}
= 1422i\frac{14}{2\sqrt{2}i} = 72i×ii\frac{7}{\sqrt{2}i} \times \frac{i}{i} = 7i2i2\frac{7i}{\sqrt{2} i^2} = 7i2(−1)\frac{7i}{\sqrt{2}(-1)} = −7i2(−1)\frac{-7i}{\sqrt{2}(-1)} = −7i2\frac{-7i}{\sqrt{2}} = −7×22×2i\frac{-7 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} i
(By rationalising the denominator)
= −72i2\frac{-7\sqrt{2}i}{2} = 0 + i (−722)\left(\frac{-7\sqrt{2}}{2}\right)
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