# Need help with complex numbers identity (1 Viewer)

#### Alasdair09

##### New Member
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

##### 617 pages
One way is by similar triangles.

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

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

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

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

and we are done.

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

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#### Trebla

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

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

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

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

Similarly

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

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

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

#### Alasdair09

##### New Member
One way is by similar triangles.

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

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

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

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

and we are done.

As a bonus, this doesn't require restrictions on $\bg_white \arg z$ 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

##### π
Moderator
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

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

#### HeroWise

##### Active Member
Eulers form is for boomers, change my mind. jkjkjkj

#### Drdusk

##### π
Moderator
Eulers form is for boomers, change my mind. jkjkjkj