Question

Find conjugate and modulus of the 5+4i / 2-3i
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Answer 1

$\sf \frac{5+4i}{2-3i}\\\\\sf </p><p>Rationalizing, \:\frac{5+4i}{2-3i} =Z\\\\\sf </p><p>= \frac{5+4i}{2-3i} x \frac{2+3i}{2+3i}\\\\\sf </p><p>=\frac{(5+4i)(2+3i)}{(2-3i)(2+3i)} \\\\\sf </p><p>=\frac{(5(2+3i)+4i(2+3i)}{(2^2-{3i} ^2)}\\\\\sf </p><p>=\frac{(10+15i+8i+12i^2)}{(4-9i^2)} \\\\\sf </p><p>=\frac{(10+23i+12i^2)}{(4-9i^2)} \\\\\sf </p><p>Putting\: i^2 =-1\\\\\sf </p><p>We \:get, \\\\\sf </p><p>=\frac{(23i-2)}{(4+9)}\\\\\sf </p><p>=\frac{(23i-2)}{(13)}\\\\\sf </p><p>=\frac{(-2+23i)}{(13)}\\\\\sf </p><p>= \frac{-2}{13} + \frac{23}{13}x \:i\\ \\\sf </p><p>Now,\: it \:is \:in \:the\: form \:x+yi =z\\\\\sf </p><p>So, x=\frac{-2}{13} and y= \frac{23}{13}\\\\\sf </p><p>Now,\: modulus \:of \:z= |z|=\sqrt{x^2+y^2}\\\\\sf </p><p>On\:solving \:we \:get\: modulus\\\\\sf </p><p>= \sqrt\frac{533}{169}</p><p>$