# Rotation in complex plane

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• December 12th 2010, 01:43 PM
atreyyu
Rotation in complex plane
If, for complex z and w, $|z+w|=|z-w|$, show that $\arg z$ and $\arg w$ differ by $\pi /2$.

I have an idea how to solve this. Drive out $a/b=-y/x$ from the initial equation (where $z=a+bi,w=x+yi$), then do something with rotating tangents around 90 degrees blahblah. The problem is I don't really know how to put this down on paper as I get kinda lost when it comes to these funky substitutions into arctan functions. Or maybe there is another way to do this?
(We didn't do vectors yet, so I should perhaps do without them, but any hint you can drop on dot and cross products and whatnot will be appreciated.) Thanks!
• December 12th 2010, 02:44 PM
tonio
Quote:

Originally Posted by atreyyu
If, for complex z and w, $|z+w|=|z-w|$, show that $\arg z$ and $\arg w$ differ by $\pi /2$.

I have an idea how to solve this. Drive out $a/b=-y/x$ from the initial equation (where $z=a+bi,w=x+yi$), then do something with rotating tangents around 90 degrees blahblah. The problem is I don't really know how to put this down on paper as I get kinda lost when it comes to these funky substitutions into arctan functions. Or maybe there is another way to do this?
(We didn't do vectors yet, so I should perhaps do without them, but any hint you can drop on dot and cross products and whatnot will be appreciated.) Thanks!

With your symbols, $(a+x)^2+(b+y)^2=|z+w|^2=|z-w|^2=(z-x)^2+(b-y)^2\Longrightarrow ax+by=0\Longrightarrow$

$\Longrightarrow \frac{b}{a}=-\frac{x}{y}=-\frac{1}{y/x}$

Now all follows from $\arctan\omega+\arctan(1/\omega)=\frac{\pi}{2}$

Tonio
• December 12th 2010, 02:50 PM
Plato
The fact that $|z-w|=|z+w|$ means that z is equally distance from w and –w.
Thus z is on the perpendicular bisector of the line segment with endpoints w and –w. That line passes through $(0,0)$.
Argue that the two complex numbers are ‘perpendicular’, (vector wise).
• December 12th 2010, 03:15 PM
atreyyu
Quote:

Originally Posted by tonio
Now all follows from $\arctan\omega+\arctan(1/\omega)=\frac{\pi}{2}$

I've gotten to that point by still I don't quite get it where it comes from...

And "to argue [...] vector wise" - I don't really know how to do it!
• December 12th 2010, 03:31 PM
Plato
Quote:

Originally Posted by atreyyu
And "to argue [...] vector wise" - I don't really know how to do it!

It is too vague only if you did not bother to draw a diagram. Did you?
Do you understand the geometry of complex numbers?
• December 12th 2010, 07:44 PM
tonio
Quote:

Originally Posted by atreyyu
I've gotten to that point by still I don't quite get it where it comes from...

And "to argue [...] vector wise" - I don't really know how to do it!

Well, the above equality can be proved either by differentiating the LHS and checking it is zero and then evaluating at $x=1$ , or

even more basically by trigonometry: take a straight angle triangle with acute angles $\alpha,\beta$ and opposite

legs of length $a,b$ , resp. Then $\displaystyle{\tan\alpha+\tan\beta=\frac{a}{b}+\fr ac{b}{a}}$ and now just

apply $\arctan$ to both sides and remember that $\displaystyle{\alpha+\beta=\frac{\pi}{2}}$

Tonio