### Author Topic: Section 1.2 Question 18  (Read 1962 times)

#### Vedant Shah

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##### Section 1.2 Question 18
« on: September 23, 2018, 05:19:33 PM »
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!

« Last Edit: September 23, 2018, 09:10:52 PM by Victor Ivrii »

#### Victor Ivrii

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##### Re: Section 1.2 Question 18
« Reply #1 on: September 23, 2018, 09:11:34 PM »
Consider arguments of two $a,c$

#### Min Gyu Woo

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##### Re: Section 1.2 Question 18
« Reply #2 on: September 24, 2018, 02:15:00 PM »
Isn't $Re(a\overline{c}) = AC + BD$?

#### Hehree Lee

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##### Re: Section 1.2 Question 18
« Reply #3 on: October 08, 2018, 07:26:55 PM »
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.