Chapter 9.8, Problem 38E

(a)

To determine

To find the complex number that a programmer should multiply to produce the transformation.

Answer to Problem 38E

  $(x,y)=(x\cos \theta ,-y\sin \theta )$

Explanation of Solution

Given:

The programmer starts with a square with vertices at $(2,2),(-2,2),(-2,-2)$ and $(2,-2)$

Calculation:

There is a need to rotate the square by ${45}^{\circ }$ counter clockwise. The vertices of the square are $(2,2),(-2,2),(-2,-2)$ and $(2,-2)$ .

So, the programmer will use the rotation matrix $\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)$ with $\theta =\frac{\pi }{4}$ .

Therefore the transformation will be $(x,y)=(x\cos \theta ,-y\sin \theta )$

This means each of the vertex will get multiplied by $\frac{1}{\sqrt{2}}$

(a)

To determine

To find the result if the original vertices are multiplied by the square of answer to part (a).

Answer to Problem 38E

The new vertices of the square will be $(1,-1),(-1,-1),(-1,1)$ and $(1,1)$ .

Explanation of Solution

Given:

The programmer starts with a square with vertices at $(2,2),(-2,2),(-2,-2)$ and $(2,-2)$

Calculation:

There is a need to rotate the square by ${45}^{\circ }$ counter clockwise. The vertices of the square are $(2,2),(-2,2),(-2,-2)$ and $(2,-2)$ .

So, the programmer will use the rotation matrix $\left(\begin{array}{cc}\cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{array}\right)$ with $\theta =\frac{\pi }{4}$ .

This means each of the vertex will get multiplied by $\frac{1}{\sqrt{2}}$

Now if we do the square of it, $\frac{1}{\sqrt{2}}$ will becomes $\frac{1}{2}$ , this means all the points will be half of the all the given points, this means the new vertices of the square will be $(1,-1),(-1,-1),(-1,1)$ and $(1,1)$ .