Question

Find the addictive inverse of (3-√-16)/(2-√-9).

This question is given in Elements of Mathematics Class 11 Exercise 5.2 Complex Numbers. Please solve it as soon as possible.
$\frac{3 - \sqrt{ - 6} }{2 - \sqrt{ - 9} }$
​

Answer 1

SOLUTION

TO DETERMINE

The additive inverse of

$\displaystyle \sf{ \frac{3 - \sqrt{ - 16} }{2 - \sqrt{ - 9} } }$

EVALUATION

We have to find the additive inverse of

$\displaystyle \sf{ \frac{3 - \sqrt{ - 16} }{2 - \sqrt{ - 9} } }$

We first simplify the expression

$\displaystyle \sf{ \frac{3 - \sqrt{ - 16} }{2 - \sqrt{ - 9} } }$

$\displaystyle \sf{ = \frac{3 - 4i }{2 - 3i } }$

$\displaystyle \sf{ = \frac{(3 - 4i)(2 + 3i) }{(2 + 3i)(2 - 3i) } }$

$\displaystyle \sf{ = \frac{6 + 9i - 8i - 12 {i}^{2} }{ {2}^{2} - 9 {i}^{2} } }$

$\displaystyle \sf{ = \frac{6 + 9i - 8i + 12 }{4 + 9 } }$

$\displaystyle \sf{ = \frac{18 + i}{13 } }$

Hence the required additive inverse

$\displaystyle \sf{ = - \bigg(\frac{18 + i}{13 } \bigg) }$

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