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What is the conjugate of $\; \large\frac{\sqrt{5+12 i } +\sqrt{5-12 i } } {\sqrt{5+12 i } -\sqrt{5-12 i } }$

$(a)\;0 + \large\frac{3}{2} i\qquad(b)\;1 - \large\frac{\sqrt{7}}{2} i\qquad(c)\;1 - \large\frac{\sqrt{5}}{16} i\qquad(d)\;0 - \large\frac{\sqrt{5}}{2} i$

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1 Answer

Answer : $\;0 + \large\frac{3}{2} i$
Explanation :
Let $\; z= \large\frac{\sqrt{5+12 i } +\sqrt{5-12 i } } {\sqrt{5+12 i } -\sqrt{5-12 i } } \times \large\frac{\sqrt{5+12 i } +\sqrt{5-12 i } } {\sqrt{5+12 i } +\sqrt{5-12 i } }$
$= \large\frac{5+12i +5-12i + 2 \sqrt{25+144}}{5+12i + 5+12i}$
$= \large\frac{3}{2i} =-\large\frac{3i}{2} = 0 - \large\frac{3}{2} i$
Therefore , The conjugate of $\;z = 0+ \large\frac{3}{2} i$
answered Apr 18, 2014 by yamini.v
 

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