Section 1.2 Question 18

[Vedant Shah]

I'm struggling with this question, and I was hoping someone could help me out: $\renewcommand{\Re}{\operatorname{Re}} \renewcommand{\Im}{\operatorname{Im}}$


Show that two lines $\Re(a+ib)=0$ and $\Re(c+id)=0$ are perpendicular  $ \iff \Re(a \bar{c}) = 0$
From section 1.2: Let $a = A+iB$ and $c= C+iD$. Then the lines are $Ax-By+\Re(b)=0$ and $Cx-Dy+\Re(d)=0$
Setting the slope of the first equal to the negative reciprocal of the other I get: $\frac{A}{B} = - \frac{D}{C} \iff AC=-BD$
Finally, $\Re(a \bar{c}) = AC-BD= 2AC$

How do I proceed?

Thanks!

[Victor Ivrii]

Consider arguments of two $a,c$

[Min Gyu Woo]

Isn't $Re(a\overline{c}) = AC + BD$?

[Hehree Lee]

Sorry, a little late, but I personally used another form of the line equation, Re[(m + 1)z + b], to relate the two perpendicular lines and solve the problem.