Solveeit Logo
Login

Question

Question: The center of a square ABCD is at \(z = 0\) . A is \({z_1}\) . Then the centroid of triangle ABC is ...

The center of a square ABCD is at z=0z = 0 . A is z1{z_1} . Then the centroid of triangle ABC is
A. z1(cos⁡π±isin⁡π){z_1}(\cos \pi \pm i\sin \pi )
B. z13(cos⁡π±isin⁡π)\dfrac{{{z_1}}}{3}(\cos \pi \pm i\sin \pi )
C. z1(cos⁡π2±isin⁡π2){z_1}\left( {\cos \dfrac{\pi }{2} \pm i\sin \dfrac{\pi }{2}} \right)
D. z13(cos⁡π2±isin⁡π2)\dfrac{{{z_1}}}{3}\left( {\cos \dfrac{\pi }{2} \pm i\sin \dfrac{\pi }{2}} \right)

Explanation

Solution

First we’ll find the coordinate of vertices of triangle ABC using the rotation formula of the complex number, after that, we’ll use the property that centroid whose formula is given by the average of the coordinates of the vertices of a triangle. Using this we’ll find the coordinate for centroid but the order of square ABCD can be clockwise or anticlockwise so we’ll consider both the cases.

Complete step by step answer:

Given data: Center of square ABCD is at z=0z = 0 and A is z1{z_1}.
Now we know that the center is z=0z = 0, we can say that z is the intersection point and midpoint of both the diagonals as diagonal of a square bisect each other.

From the figure, we can say that Z is the midpoint of side AC as AC is the diagonal of square ABCD
Therefore we can say that C will be−z1 - {z_1} as C is equidistant from the origin as A and is exactly opposite to A.
We know that in complex number when any point let zk{z_k} is rotated through an angle θ\theta with-respect-to origin in the anticlockwise direction then the new point will bezkeiθ{z_k}{e^{i\theta }}

Similarly, we can say that B will be z1e−iπ2{z_1}{e^{ - i\dfrac{\pi }{2}}}since the origin is the center of square ABCD and the diagonals are perpendicular to each other.
Now we know that coordinate of a centroid is given by the average of the coordinates of the vertices of the triangle.
Therefore in triangle ABC coordinate of centroid will be z1+z1e−iπ2−z13\dfrac{{{z_1} + {z_1}{e^{ - i\dfrac{\pi }{2}}} - {z_1}}}{3}
=z13e−iπ2= \dfrac{{{z_1}}}{3}{e^{ - i\dfrac{\pi }{2}}}
Now, from the above case, we can say that centroid of the triangle ABC is z13e−iπ2\dfrac{{{z_1}}}{3}{e^{ - i\dfrac{\pi }{2}}}
In the above case, the vertices ABCD are in clock wise, but if the ABCD make an anti-clockwise sequence i.e.

Now in this case B will be z1eiπ2{z_1}{e^{i\dfrac{\pi }{2}}} resulting the centroid to be z13eiπ2\dfrac{{{z_1}}}{3}{e^{i\dfrac{\pi }{2}}}.
From the above results, we can say that centroid of the triangle ABC is z13e±iπ2\dfrac{{{z_1}}}{3}{e^{ \pm i\dfrac{\pi }{2}}}

Using eiθ=cos⁡θ+isin⁡θ{e^{i\theta }} = \cos \theta + i\sin \theta
⇒z13e±iπ2=z13(cos⁡(±π2)+isin⁡(±π2))\Rightarrow \dfrac{{{z_1}}}{3}{e^{ \pm i\dfrac{\pi }{2}}} = \dfrac{{{z_1}}}{3}\left( {\cos \left( { \pm \dfrac{\pi }{2}} \right) + i\sin \left( { \pm \dfrac{\pi }{2}} \right)} \right)
Using cos⁡θ=cos⁡(−θ)\cos \theta = \cos \left( { - \theta } \right)andsin⁡(−θ)=−sin⁡θ\sin \left( { - \theta } \right) = - \sin \theta
⇒z13e±iπ2=z13(cos⁡π2±isin⁡π2)\Rightarrow \dfrac{{{z_1}}}{3}{e^{ \pm i\dfrac{\pi }{2}}} = \dfrac{{{z_1}}}{3}\left( {\cos \dfrac{\pi }{2} \pm i\sin \dfrac{\pi }{2}} \right)
Option(D) is correct.

Note: In the above solution we stated that when a point let zk{z_k} is rotated through an angle θ\theta with-respect-to origin in the anticlockwise direction then the new point will bezkeiθ{z_k}{e^{i\theta }}, we apply this formula only when the point is measured with-respect-to origin and both the resultant complex number and the original number is equidistant from the origin and if not then we use the general formula.