## Engage NY Eureka Math Precalculus Module 1 Mid Module Assessment Answer Key

### Eureka Math Precalculus Module 1 Mid Module Assessment Task Answer Key

Question 1.

Given z=3-4i and w=-1+5i:

a. Find the distance between z and w.

Answer:

Distance = \(\sqrt{(3-(-1))^{2}-(-1)+((-4)-5)^{2}}\)

=\(\sqrt{4^{2}+(-9)^{2}}\)

= \(\sqrt{16+81}\)

= \(\sqrt{97}\)

b. Find the midpoint of the segment joining z and w.

Answer:

Midpoint =\(\frac{3+(-1)}{2}\) + \(\frac{(-4)+5}{2} i\)

=\(\frac{2}{2}\) +\(\frac{1}{2} i\)

=1+\(\frac{1}{2}\)i

Question 2.

Let z_{1}=2-2i and z_{2}=(1-i)+\(\sqrt{3}\)(1+i).

a. What is the modulus and argument of z_{1}?

Answer:

b. Write z_{1} in polar form. Explain why the polar and rectangular forms of a given complex number represent the same number.

z_{1} = 2\(\sqrt{2}\) [cos((-\(\frac{\pi}{4}\))+i sin((-\(\frac{\pi}{4}\))]

The modulus represents the distance from the origin to the point. The degree of rotation is the angle from the x-axis. When the polar form is expanded, the result is the rectangular form of a complex number.

c. Find a complex number w, written in the form w=a+ib, such that wz_{1}=z_{2}.

Answer:

d. What is the modulus and argument of w?

Answer:

|w| = \(\sqrt{\left(\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}}\) = \(\sqrt{\frac{1}{4}+\frac{3}{4}}\) = 1

arg (w )=tan^{-1}(\(\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}\))=\(\frac{\pi}{3}\)

e. Write w in polar form.

Answer:

w=1 [cos(\(\frac{\pi}{3}\))+i sin (\(\frac{\pi}{3}\)) ]

f. When the points z_{1} and z_{2} are plotted in the complex plane, explain why the angle between z_{1} and z_{2} measures arg(w).

Answer:

Since z_{2} = wz_{1}, then z_{2}is the transformation of z_{1} rotated counterclockwise by arg (w ), which is \(\frac{\pi}{3}\).

g. What type of triangle is formed by the origin and the two points represented by the complex numbers z_{1} and z_{2}? Explain how you know.

Answer:

Since |w | = 1 and arg (w ) = \(\frac{\pi}{3}\), the triangle formed by the origin and the points representing z_{1} and z_{2}will be equilateral. All of the angles are 60° in this triangle.

h. Find the complex number, v, closest to the origin that lies on the line segment connecting z_{1} and z_{2}. Write v in rectangular form.

Answer:

The point that represents v is the midpoint of the segment connecting z_{1} and z_{2}since it must be on the perpendicular bisector of the triangle with vertex at the origin.

To find the midpoint, average z_{1} and z_{2}.

Question 3.

Let z be the complex number 2+3i lying in the complex plane.

a. What is the conjugate of z? Explain how it is related geometrically to z.

Answer:

\(\overline{\boldsymbol{Z}}\)= 2-3i

This number is the conjugate of z and the reflection of z across the horizontal axis.

b. Write down the complex number that is the reflection of z across the vertical axis. Explain how you determined your answer.

Answer:

This number is -2 + 3i . The real coordinate has the opposite sign, but the imaginary part keeps the same sign. This means a reflection across the imaginary (vertical) axis.

Let m be the line through the origin of slope \(\frac{1}{2}\) in the complex plane.

c. Write down a complex number, w, of modulus 1 that lies on m in the first quadrant in rectangular form.

Answer:

Because the slope of m is \(\frac{1}{2}\), the argument of w is tan^{-1}(\(\frac{1}{2}\)).

Using the polar form of w ,

w = 1 [cos(tan^{-1}(\(\frac{1}{2}\)) ) + isin (tan^{-1}(\(\frac{1}{2}\)) )]. From the triangle shown below, w = \(\frac{2}{\sqrt{5}}\) + \(\frac{1}{\sqrt{5}}\)i .

d. What is the modulus of wz?

Answer:

The modulus of wz is \(\sqrt{13}\).

e. Explain the relationship between wz and z. First, use the properties of modulus to answer this question, and then give an explanation involving transformations.

Answer:

The modulus of wz is \(\sqrt{13}\) and is the same as |z |.

Using the properties of modulus,

|wz | = |w | × |z | = 1 × |z | = 1 × \(\sqrt{2^{2}+3^{2}}\) = \(\sqrt{13}\).

Geometrically, multiplying by w will rotate z by the arg (w ) and dilate z by |w |. Since |w | = 1 , the transformation is a rotation only, so both w and z are the same distance from the origin.

f. When asked,

“What is the argument of \(\frac{1}{w}\)z?”

Paul gave the answer arctan(\(\frac{3}{2}\))-arctan(\(\frac{1}{2}\)), which he then computed to two decimal places. Provide a geometric explanation that yields Paul’s answer.

Answer:

The product \(\frac{1}{w}\)z would result in a clockwise rotation of z by the arg (w ). There would be no dilation since |w | = 1 .

arg (z ) = tan^{-1}(\(\frac{3}{2}\))

arg (w ) = tan^{-1}(\(\frac{1}{2}\))

arg (\(\frac{1}{w}\)z ) = tan^{-1}(\(\frac{3}{2}\)) – tan^{-1}(\(\frac{1}{2}\))

g. When asked,

“What is the argument of \(\frac{1}{w}\)z?”

Mable did the complex number arithmetic and computed z÷w. She then gave an answer in the form arctan(\(\frac{a}{b}\)) for some fraction \(\frac{a}{b}\). What fraction did Mable find? Up to two decimal places, is Mable’s final answer the same as Paul’s?

Answer:

Comparing these angles shows they are the same.

tan^{-1}(\(\frac{4}{7}\))≈0.52

tan^{-1}(\(\frac{3}{2}\))- tan^{-1}(\(\frac{1}{2}\))≈0.52