Question

The centre of a square ABCD is at $z=0$. A is $z_1$. Then the centroid of triangle ABC is

• A. $z_1(cos\pi \pm i\sin \pi )$
• B. $\frac{z_1}{3}(\cos \pi \pm i\sin \pi )$
• C. $z_1(\cos \frac{\pi }{2}\pm i\sin \frac{\pi }{2})$
• D. $\frac{z_1}{3}(\cos \frac{\pi }{2}\pm i\sin \frac{\pi }{2})$

Answer 1

The correct option is D $\frac{z_1}{3}(\cos \frac{\pi }{2}\pm i\sin \frac{\pi }{2})$

Let $z=r{e}^{i\theta }$
Hence, B point will be represented by $z=r{e}^{i(\theta +\frac{\pi }{2})}$
The point C will be represented by $z=r{e}^{i(\theta +\pi )}$
Hence, their centroid is represented by=$\frac{r{e}^{i\left(\theta \right)}+r{e}^{i(\theta +\frac{\pi }{2})}+r{e}^{i(\theta +\pi )}}{3}$
$=\frac{r{e}^{i\left(\theta \right)}(1+r{e}^{i\left(\frac{\pi }{2}\right)}+r{e}^{i\left(\pi \right)})}{3}$
$=\frac{z(1+(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2})+(\cos \pi +i\sin \pi ))}{3}$
$=\frac{z(1+(0+i)+(-1+0\left)\right)}{3}$
$=\frac{z\left(i\right)}{3}$
$=\frac{z\left(\cos \frac{\pi }{2}+i\sin \frac{\pi }{2}\right)}{3}$