Complex numbers question (2 Viewers)
- Thread starter aonin
- Start date
Hermes1
Banned
- Joined
- Oct 4, 2010
- Messages
- 1,282
- Gender
- Male
- HSC
- 2011
the diff of two squares is wrongHey guys I'm stumped on this question, unfortunately there weren't any solutions provided
View attachment 22973
thecame from the fact that
forms an equilateral triangle with a side
and a top side of length
, hence since
is this reasoning right?
Thanks
K
khorne
Guest
I dare say from my diagram I can't see it working.
Hermes1
Banned
- Joined
- Oct 4, 2010
- Messages
- 1,282
- Gender
- Male
- HSC
- 2011
agree with this. i hav seen questions like this b4 and ur question doesnt seem to work
K
khorne
Guest
Draw the circles, |z|=a and |z^2| = a^2.
Construct the triangles, they are similar. There is one triangle in the |z|=a circle, across the diameter. It has angles theta (at z+a) and 90-theta (at z-a). Do the same for |z^2| = a^2. Those two triangles are similar, so arg(z^2 - a^2) = 90+theta. Thus, in the triangle made by z^2 -a, z^2 -a^2 and a^2 -a, we get one angle being 90-theta. This means the other (i.e arg (z^2 -a)) cannot be 90+theta, otherwise we'd have a wonky triangle. This could only work for say, |z|=1
Construct the triangles, they are similar. There is one triangle in the |z|=a circle, across the diameter. It has angles theta (at z+a) and 90-theta (at z-a). Do the same for |z^2| = a^2. Those two triangles are similar, so arg(z^2 - a^2) = 90+theta. Thus, in the triangle made by z^2 -a, z^2 -a^2 and a^2 -a, we get one angle being 90-theta. This means the other (i.e arg (z^2 -a)) cannot be 90+theta, otherwise we'd have a wonky triangle. This could only work for say, |z|=1
Isosceles rather than necessarily an equilateral triangle, I think.\
there is an equilateral triangle created due to the vector of
Hermes1
Banned
- Joined
- Oct 4, 2010
- Messages
- 1,282
- Gender
- Male
- HSC
- 2011
Circle geometry is what i used to solve this.Thanks for the tip! Heres how I finally did it
=\arg(z+a)(z-a)=\arg(z+a)+\arg(z-a)\\arg(z-a)=\left(\frac{\pi-\theta}{2}\right)+\theta=\frac{\pi}{2}+\frac{\theta}{2}\\arg(z+a)=\frac{\theta}{2}\quad\text{as diagonals of rhombus bisect angles in corner}\\\therefore \arg(z+a)(z-a)=\frac{\pi}{2}+\frac{\theta}{2}+\frac{\theta}{2}=\frac{\pi}{2}+\theta)