Answers: 1
![\bf \cfrac{\sqrt{-49}}{(7-2i)-(4-9i)}\implies \cfrac{\sqrt{-1\cdot 49}}{(7-2i)-4+9i}\implies \cfrac{\sqrt{-1}\cdot \sqrt{49}}{3+7i} \\\\\\ \stackrel{\textit{multiplying top/bottom by the conjugate of the denominator }(3-7i)}{\cfrac{7i}{3+7i}\cdot \cfrac{3-7i}{3-7i}\implies \cfrac{7i(3-7i)}{\stackrel{\textit{difference of squares}}{(3+7i)(3-7i)}}\implies \cfrac{21i-(7i)^2}{[3^2-(7i)^2]}}](/tpl/images/0052/4554/fc04d.png)
![\bf \cfrac{21i-(7^2i^2)}{[9-(7^2i^2)]}\implies \cfrac{21i-(49(-1))}{[9-(49(-1))]}\implies \cfrac{21i+49}{9+49}\implies \cfrac{49+21i}{58}](/tpl/images/0052/4554/e17ba.png)
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Rationalize the denominator of: [tex] \frac{ \sqrt{ - 49} }{(7 - 2i) - (4 - 9i)} [/tex]a. [tex] \fra...
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