0
$\begingroup$

Let $O, Z_1, Z_2, Z_3$ be the respective vertices of a rhombus such that $O$ is the origin, $|Z_1|=|Z_2|=4$ and $|Z_3|=6$. What is $\arg(Z_3)$, if $\arg(Z_2-Z_1)=\dfrac{\pi}{3}$?

I tried to solve this question by taking $Z_1=2(\cos a+i\sin a)$ and $Z_2=2(\cos b+i\sin b)$.Then $\arg(Z_2-Z_1)=\arg(2((\cos a-\cos b)+i(\sin a-\sin b))$. I could not proceed after this. I also tried to use cosine formula but it too did not help me. Any ideas to go ahead would be highly appreciated.

Thanks.

$\endgroup$
1
$\begingroup$

Since a coordinate system hasn't already been imposed, let me just consider that $A(Z_1)$ lies on the negative $x$-axis, and $B(Z_2)$ lies in the second quadrant. Then $C(Z_3)$ also lies in the second quadrant. We're given that $\angle BAO =\pi/3$. In a rhombus, the diagonals bisect the internal angles. Hence $\angle CAO =2\pi/3$. Also, since $CA \ || \ BO$, we have $\angle BOD =2\pi/3$. It follows that $\angle BOA = \pi/3$. And since diagonals bisect internal angles, $\angle BOC = \pi/6$. Thus, $\arg(Z_3) = \angle BOC + \angle BOD = 5\pi/6$. Now, the rhombus could also be reflected about the origin to obtain $\arg(Z_3) = -\pi/6$. Any other orientation of the rhombus violates $\arg(Z_2-Z_1) =\pi/3$.

EDIT: $D$ is an arbitrary point on the positive $x$-axis.

enter image description here

$\endgroup$
1
$\begingroup$

Let $Z_1=4\operatorname{cis} \theta$ and $Z_2=4\operatorname{cis} \phi$.

By parallelogram law,

$$Z_3=Z_1+Z_2=8\cos \frac{\theta-\phi}{2} \operatorname{cis} \frac{\theta+\phi}{2}$$

$$\cos \frac{\theta-\phi}{2}=\frac{3}{4}$$ which is consistent with cosine law.

Also, $$\arg Z_3 = \frac{\theta+\phi}{2}$$

Now \begin{align*} Z_2-Z_1 &= 4(\operatorname{cis} \theta-\operatorname{cis} \phi) \\ &= 2\sin \frac{\theta-\phi}{2} \left( -\sin \frac{\theta+\phi}{2}+i\cos \frac{\theta+\phi}{2} \right) \\ &= 2\sin \frac{\theta-\phi}{2} \operatorname{cis} \frac{\theta+\phi+\pi}{2} \\ \end{align*}

That means

$$Z_1 Z_2 \perp OZ_3$$

Therefore

$$\arg Z_3=\frac{5\pi}{6} \quad \text{or} \quad \frac{11\pi}{6}$$

If I was the teacher, I just requires the materials in the yellow boxes only.

$\endgroup$
1
$\begingroup$

Let $z_1=4(\cos\alpha+i\sin\alpha)$, $z_2=4(\cos\beta+i\sin\beta)$ then $$z_2-z_1=4\Big[\cos\beta-\cos\alpha+i\sin\beta-i\sin\alpha\Big]$$ thus $$\tan\frac{\pi}{3}=\tan\arg(z_2-z_1)=\frac{\sin\beta-\sin\alpha}{\cos\beta-\cos\alpha}=-\cot\frac{\alpha+\beta}{2}=-\tan\left(\frac{\pi}{2}-\frac{\alpha+\beta}{2}\right)$$ or $$\frac{\pi}{3}=-\frac{\pi}{2}+\frac{\alpha+\beta}{2}~~~~;~~~~\frac{\pi}{3}=\pi-\frac{\pi}{2}+\frac{\alpha+\beta}{2}$$ With a simple geometric consideration, we see that rhombus diagonals are perpendicular, and $z_3$ is the bisector of $z_1$ and ‎$‎z_2$‎, so $$\arg z_3=\frac{\alpha+\beta}{2}=\color{blue}{\dfrac{5\pi}{6}, \dfrac{-\pi}{6}}$$

$\endgroup$
  • $\begingroup$ How can you say that $z_2=4\exp(i(\pi/3+\alpha))$? That is wrong. $\endgroup$ – Aretino Feb 16 '17 at 14:30
  • $\begingroup$ And in your last equation, how can the product of three factors, all of them not vanishing, be zero? That equation has no solutions. $\endgroup$ – Aretino Feb 16 '17 at 14:32
  • $\begingroup$ The product of two complex numbers cannot vanish, if none of the factors vanishes. $\endgroup$ – Aretino Feb 16 '17 at 14:36
  • $\begingroup$ @Aretino What is a strange mistake. you are right. will edit. $\endgroup$ – Nosrati Feb 16 '17 at 14:41
0
$\begingroup$

According to your data, rhombus diagonal $Z_1Z_2$ forms an angle of $\pi/3$ with the real axis. It follows that the other diagonal $OZ_3$, which is perpendicular to $Z_1Z_2$, forms an angle of $\pi/3\pm\pi/2$ with the real axis: $\arg Z_3=\pi/3\pm\pi/2$.

$\endgroup$
0
$\begingroup$

Just draw a picture and use simple geometry. $Arg(Z_3) = 2 \pi - \frac{\pi}{6}$ when $OZ_1Z_3Z_2$ is counter-clockwise oriented or $Arg(Z_3) = \pi - \frac{\pi}{6}$ when $OZ_1Z_3Z_2$ is clockwise oriented

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.