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The equation of the line through $2 + 3i$ and $0$ can be written as $az + b \overline{z} = 0$for some complex numbers $a$ and $b$. Find the quotient $b/a$ in rectangular form.

Jan 18, 2019

#1
+23137
+9

The equation of the line through $2 + 3i$ and $0$ can be written as $az + b \overline{z} = 0$for some complex numbers $a$ and $b$.

Find the quotient $b/a$ in rectangular form.

$$\begin{array}{|rcll|} \hline az + b \overline{z} &=& 0 \\ &&\boxed{ z = 2+3i} \\ && \boxed{\overline{z} = 2-3i} \\ a(2+3i) + b (2-3i) &=& 0 \\ b (2-3i) &=& -a(2+3i) \\\\ \dfrac{b}{a}&=& \dfrac{-(2+3i)} {(2-3i)} \\\\ \dfrac{b}{a}&=& \dfrac{-(2+3i)} {(2-3i)}\cdot\dfrac{(2+3i)} {(2+3i)} \\\\ \dfrac{b}{a}&=& \dfrac{-(2+3i)(2+3i)} {(2-3i)(2+3i)} \\\\ \dfrac{b}{a}&=& \dfrac{-(4+12i+9i^2)} {4-9i^2} \quad & | \quad i^2=-1 \\\\ \dfrac{b}{a}&=& \dfrac{-(4+12i-9)} {4+9} \\\\ \dfrac{b}{a}&=& \dfrac{-(-5+12i)} {13} \\\\ \dfrac{b}{a}&=& \dfrac{5-12i} {13} \\\\ \mathbf{\dfrac{b}{a}} &\mathbf{=}& \mathbf{\dfrac{5} {13} -\dfrac{12} {13}i} \\ \hline \end{array}$$

Jan 18, 2019