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If $A(z_1)$ and $(z_2)$ are two points in argand(complex) plane such that $$\frac{z_1}{z_2}+\frac{\overline{z_1}}{\overline{z_2}}=2$$. Find the value of $\angle ABO$ where $O$ is origin.

Using given condition, I found that Real part of $\frac{z_1}{z_2}=1$ but I am not able to use this to find $\angle ABO$. Could someone help me with this?

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If we write $z_1=\rho_1 e^{i\theta_1}$ and $z_2=\rho_2 e^{i\theta_2}$ (with $z_1\ne z_2$, if $z_1=z_2$ we have trivially $\beta=0$), then from the equation we get:

$${\rho_1 e^{i\theta_1}\over\rho_2 e^{i\theta_2}}+{\rho_1 e^{-i\theta_1}\over\rho_2 e^{-i\theta_2}}=2$$ $$e^{i(\theta_1-\theta_2)}+e^{-i(\theta_1-\theta_2)}=2{\rho_2\over\rho_1}$$ $${e^{i(\theta_1-\theta_2)}+e^{-i(\theta_1-\theta_2)}\over2}={\rho_2\over\rho_1}$$ $$\cos(\theta_1-\theta_2)={\rho_2\over\rho_1}$$

so we get $\rho_2\le\rho_1$.

Now we can apply the law of sines to $\triangle ABO$:

$${\sin\beta\over\rho_1}={\sin(\theta_1-\theta_2)\over\overline {AB}}\longrightarrow\sin\beta=\rho_1{\sin(\theta_1-\theta_2)\over\overline {AB}}$$

where $\beta=\angle ABO$.

We know that:

$$\sin(\theta_1-\theta_2)=\sqrt{1-\cos^2(\theta_1-\theta_2)}={\sqrt{\rho_1^2-\rho_2^2}\over\rho_1}$$

and applying the cosine rule to $\triangle ABO$:

$$\overline{AB}=\sqrt{\rho_1^2+\rho_2^2-2\rho_1\rho_2\cos(\theta_1-\theta_2)}=\sqrt{\rho_1^2+\rho_2^2-2\rho_1\rho_2{\rho_2\over\rho_1}}=\sqrt{\rho_1^2-\rho_2^2}$$

hence we get: $\sin\beta=1\longrightarrow\beta={\pi\over2}$. The angle $\beta\equiv\angle ABO$ is always ${\pi\over2}$.

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Let $z_1 / z_2=a+bi$ with $a,b \in \mathbb{R}$. Then $\bar z_1 / \bar z_2=a-bi$ and the given condition gives $(a+bi)+(a-bi) = 2 a = 2 \iff a = 1 \iff z_1/z_2 = 1 + bi$.

The angle $\angle ABO = \arg((z_1-z_2) / z_2)=\arg(z_1/z_2-1)=\arg(1+bi-1)=\arg(bi) = \pm \pi / 2$. Ignoring orientation, $\angle ABO = \pi / 2\,$.

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If we write $z_1=x+iy$ and $z_2=u+iv$ then the given condition reduces to the relation $$xu+yv=u^2+v^2$$

The required angle is given by $$\arg\frac {z_2}{z_2-z_1}$$

This simplifies to become $$\arg\frac{iyu-ivx}{(u-x)^2+(v-y)^2}$$

Therefore, provided that $A$ and $B$ are not identical or collinear with the origin, the required angle is $90^o$

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Assume the points $O,A,B$ are distinct.

Let $A = z_1 = s + ti$ and $B = z_2 = u + vi$. Then $$\frac{z_1}{z_2} + \frac{\overline{z_1}}{\overline{z_2}} =2 $$ $$\Rightarrow \frac{s + ti}{u + vi} + \frac{s - ti}{u - vi} = 2$$ Combining and simplifying the above equation yields

$$su + tv = u^2 + v^2$$

$$\Rightarrow u^2 + v^2 - su - tv = 0$$

$$\Rightarrow (u - s/2)^2 + (v - t/2)^2 = (s^2 + t^2)/4$$

$$\Rightarrow |z_2 - z_1/2| = |z_1|/2$$

Then $0,z_1,z_2$ are equidistant from $z_1/2$, hence, since $O,A,B$ are distinct, it follows that $B$ is on the semicircle with diameter $OA$. Therefore angle $ABO$ is a right angle.

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