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.

5 Answers 5

0

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

1

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.

1

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 $$\arg z_3=\frac{\alpha+\beta}{2}=\color{blue}{\dfrac{5\pi}{6}, \dfrac{-\pi}{6}}$$

  • 0
    How can you say that $z_2=4\exp(i(\pi/3+\alpha))$? That is wrong. – 2017-02-16
  • 0
    And in your last equation, how can the product of three factors, all of them not vanishing, be zero? That equation has no solutions. – 2017-02-16
  • 0
    The product of two complex numbers cannot vanish, if none of the factors vanishes. – 2017-02-16
  • 0
    @Aretino What is a strange mistake. you are right. will edit. – 2017-02-16
1

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

0

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: $\arg Z_3=\pi/3\pm\pi/2$.