Need help with complex numbers identity (1 Viewer)

Alasdair09 · Dec 9, 2019

Question: "The points A, B, C and O represent the numbers z, 1/z, 1 and 0 respectively. Given that 0<argz<pi/2 prove that the angle <OAC = <OCB"
Please try this question and see if you can find an elegant proof thank you.

fan96 · Dec 9, 2019

One way is by similar triangles.

We suppose that $A,B,C,O$ are distinct points (otherwise the question isn't very well defined).
In particular, $z \ne 0, 1$ . Then,

$\frac{OC}{OA} = \frac{|1|}{|z|},$

$\frac{BC}{CA} = \frac{|1-1/z|}{|z-1|} =\frac{|1/z||z-1|}{|z-1|},$

$\frac{OB}{OC} = \frac{|1/z|}{|1|},$

and we are done.

As a bonus, this doesn't require restrictions on $\arg z$ and you also get equalities for the other two angles of the triangles.

Trebla · Dec 10, 2019

Not as elegant but a more intuitive approach is to use the vector rotations.

$\text{Let }\angle OAC = \alpha\:\text{and}\:\angle OCB = \beta\:\text{then}$

$\dfrac{z}{|\overrightarrow{AO}|}e^{i\alpha}=\dfrac{1-z}{|\overrightarrow{AC}|}$

$e^{i\alpha} = \dfrac{|z|(1-z)}{z|1-z|}$

Similarly

$\dfrac{1}{|\overrightarrow{CO}|}e^{i\beta}=\dfrac{\dfrac{1}{z}-1}{|\overrightarrow{CB}|}$

$e^{i\beta} = \dfrac{|z|(1-z)}{z|1-z|}$

Since the angles lie between zero and 180 degrees then they must be equal.

Alasdair09 · Dec 10, 2019

fan96 said:
One way is by similar triangles.

We suppose that $A,B,C,O$ are distinct points (otherwise the question isn't very well defined).
In particular, $z \ne 0, 1$ . Then,

$\frac{OC}{OA} = \frac{|1|}{|z|},$

$\frac{BC}{CA} = \frac{|1-1/z|}{|z-1|} =\frac{|1/z||z-1|}{|z-1|},$

$\frac{OB}{OC} = \frac{|1/z|}{|1|},$

and we are done.

As a bonus, this doesn't require restrictions on $\arg z$ and you also get equalities for the other two angles of the triangles.

Thanks for this solution! Although the only thing I do not understand is how you go from |1-1/z|/|z-1| to |1/z||z-1|/|z-1|

Drdusk · Dec 10, 2019

Alasdair09 said:
Thanks for this solution! Although the only thing I do not understand is how you go from |1-1/z|/|z-1| to |1/z||z-1|/|z-1|

He pulled out a factor of 1/z from the numerator as 1/z * z = 1 and 1/z * 1 = 1/z so like

$\dfrac{|1-1/z|}{|z-1|} = \dfrac{|1/z(z-1)|}{|z-1|}$

HeroWise · Dec 10, 2019

Eulers form is for boomers, change my mind. jkjkjkj

Drdusk · Dec 10, 2019

HeroWise said:
Eulers form is for boomers, change my mind. jkjkjkj

Well your not wrong...

Need help with complex numbers identity (1 Viewer)

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