We are asked to consider a situation where a preimage is rotated $360$ degrees about the origin. Let's consider an example figure.

Let's rotate this figure

36 0^{\circ}

about the origin.

As we can see, the image of the rotation is exactly the same as the preimage. This is because

36 0^{\circ}

is a complete angle, so the rotation maps every point onto itself. Alternatively, we could see a rotation by

36 0^{\circ}

as a rotation by

27 0^{\circ}

followed by a rotation by

9 0^{\circ} .

27 0^{\circ} + 9 0^{\circ} = 36 0^{\circ}

A rotation by

27 0^{\circ}

maps a point with coordinates

(x, y)

onto the point with coordinates

(y, - x) .

A rotation by

9 0^{\circ}

maps a point with coordinates

(x, y)

onto the point with coordinates

(- y, x) .

Rotation by 27 0^{\circ} Rotation by 9 0^{\circ} : (x, y) \to (y, - x) : (x, y) \to (- y, x)

Let's see what happens to a point with coordinates

(a, b)

when we rotate it by

27 0^{\circ}

and then by

9 0^{\circ} .

A rotation by

27 0^{\circ}

switches the coordinates around and multiplies the second coordinate of the image by

- 1 .

Then, after the first rotation, the point is mapped onto the point

(b, - a) .

(a, b) \to (b, - a)

A rotation by

9 0^{\circ}

also swaps the coordinates around. Additionally, it multiplies the first coordinate of the image by

- 1 .

Let's see what happens when we rotate a point with the coordinates

(b, - a)

by

9 0^{\circ} .

(b, - a) \to (- (- a), b)

Note that

- (- a))

is the same as

a .

(- (- a), b) = (a, b)

This confirms that for any point, the image of a rotation by

36 0^{\circ}

about the origin is the same point as the original.

Exercise