Question

Express as a complex number in a+bi form: 2-10i/-3-7i

Answer 1

$\large\begin{array}{l} \mathsf{z=\dfrac{2-10i}{-3-7i}}\\\\\\ \textsf{Multiply and divide by the denominator's conjugate }\mathsf{(-3+7i):}\\\\ \mathsf{z=\dfrac{2-10i}{-3-7i}\cdot \dfrac{-3+7i}{-3+7i}}\\\\ \mathsf{z=\dfrac{(2-10i)\cdot (-3+7i)}{(-3-7i)\cdot (-3+7i)}} \end{array}$

$\large\begin{array}{l} \textsf{Multiply brackets out:}\\\\ \mathsf{z=\dfrac{(2-10i)\cdot (-3)+(2-10i)\cdot 7i}{(-3-7i)\cdot (-3)+(-3-7i)\cdot 7i}}\\\\ \mathsf{z=\dfrac{-6+30i+14i-70i^2}{9+\,\diagup\!\!\!\!\!\! 21i-\diagup\!\!\!\!\!\! 21i-49i^2}}\qquad\quad\textsf{(but }\mathsf{i^2=-1}\textsf{)}\\\\ \mathsf{z=\dfrac{-6+30i+14i-70\cdot (-1)}{9-49\cdot (-1)}}\\\\ \mathsf{z=\dfrac{-6+44i+70}{9+49}} \end{array}$

$\large\begin{array}{l} \mathsf{z=\dfrac{-6+70+44i}{58}}\\\\ \mathsf{z=\dfrac{64+44i}{58}}\\\\ \mathsf{z=\dfrac{\diagup\!\!\!\! 2\cdot (32+22i)}{\diagup\!\!\!\! 2\cdot 29}}\\\\ \mathsf{z=\dfrac{32+22i}{29}}\\\\\\ \textsf{Split into two fractions:}\\\\ \mathsf{z=\dfrac{32}{29}+\dfrac{22}{29}\,i} \end{array}$


$\large\begin{array}{l} \therefore~~\boxed{\begin{array}{c} \mathsf{\dfrac{2-10i}{-3-7i}=\dfrac{32}{29}+\dfrac{22}{29}\,i} \end{array}}\qquad\checkmark\\\\\\ \textsf{which already is in }\mathsf{a+bi}\textsf{ form:}\\\\ \mathsf{a=\dfrac{32}{29}}\textsf{ and }\mathsf{b=\dfrac{22}{9}\,\cdot} \end{array}$


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$\large\textsf{I hope it helps.}$