Core practice ← Back to lesson

The Complex Plane and Modulus: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Advanced (beyond the core course) Advanced. This problem set goes beyond core Algebra II. You can skip it.

0 of 10 completed

Progress saved in this browser.

Problem 1 of 10

Problem 1 A plotted marker

The figure shows the point $A$ in the complex plane. Write the complex number represented by $A$ .

The point $A$ in the complex plane.

Text description of this figure

A grid with the Real axis running from -5 to 5 and the Imaginary axis running from -5 to 5, gridlines and number labels at every integer. A single solid point, labeled A, is plotted one unit to the right of the origin and four units below it. No coordinates or complex number are printed.
1. Hint 1
  The horizontal coordinate is the real part.
2. Hint 2
  The vertical coordinate is the real coefficient multiplying $i$ .
Answer

$1-4i$ .

Full solution

The grid places $A$ one unit right and four units down from the origin.
$A=1+(-4)i$
Thus the represented number is $1-4i$ .

Answer

$1-4i$ .

Key idea

The two coordinates of a plotted complex number are its real and imaginary parts.
Problem 1: Mark complete
Problem 2 A displacement length

Find $|(-1+6i)-(-3-i)|$ exactly.
1. Hint 1
  The modulus of a difference measures the length of the displacement.
2. Hint 2
  Subtract the parts first, then apply the Pythagorean theorem.
Answer

$\sqrt{53}$ .

Full solution

The difference is $2+7i$ .
Its length is
$|2+7i|=\sqrt{2^2+7^2}$
This gives $\sqrt{53}$ , consistent with a horizontal gap of $2$ and a vertical gap of $7$ .

Answer

$\sqrt{53}$ .

Key idea

Distance between complex points is the modulus of their difference.
Problem 2: Mark complete
Problem 3 A reflected marker

The figure shows a complex number $B$ . State the coordinates of the point for $\overline B$ .

The point $B$ in the complex plane.

Text description of this figure

A grid with the Real axis running from -5 to 5 and the Imaginary axis running from -5 to 5, gridlines and number labels at every integer. A single solid point, labeled B, is plotted two units to the right of the origin and four units below it. Its conjugate is not shown.
1. Hint 1
  Conjugation reflects a point across the horizontal real axis.
2. Hint 2
  Keep the horizontal coordinate and reverse the vertical coordinate.
Answer

$(2,4)$ .

Full solution

The plotted point is $(2,-4)$ .
Conjugation changes $2-4i$ into $2+4i$ , so the reflected coordinates are $(2,4)$ .
The two points lie equally far below and above the real axis.

Answer

$(2,4)$ .

Key idea

Conjugation preserves the real coordinate and reverses the imaginary coordinate.
Problem 3: Mark complete
Problem 4 A completed route

The figure shows an arrow from $0$ to $z$ and, starting at $z$ , a copy of the arrow for $w$ . Read $z$ and $w$ , calculate $z+w$ , and give the remaining corner $w$ of the addition parallelogram.

An arrow from $0$ to $z$ , followed by a copy of the arrow for $w$ .

Text description of this figure

A grid with the Real axis from -4 to 5 and the Imaginary axis from -1 to 5, gridlines and number labels at every integer. A solid arrow starts at the origin and ends at a point labeled z, two units left and one unit up from the origin. A second, dashed arrow labeled w starts at that same point and ends at a solid point labeled E, three units right and two units up from z. No coordinate pairs are printed.
1. Hint 1
  The second arrow represents a displacement, not its endpoint coordinates.
2. Hint 2
  Subtract its starting coordinates from its ending coordinates to read $w$ .
Answer

$z=-2+i$ ; $w=3+2i$ ; $z+w=1+3i$ ; remaining corner $(3,2)$ .

Full solution

The first arrow ends at $(-2,1)$ , so $z=-2+i$ .
The second runs from $(-2,1)$ to $(1,3)$ , a displacement of $(3,2)$ , so $w=3+2i$ .
$z+w=(-2+3)+(1+2)i$
Thus $z+w=1+3i$ .
Starting the same displacement at $0$ reaches $(3,2)$ , the missing corner; either route to the final point gives the same sum.

Answer

$z=-2+i$ ; $w=3+2i$ ; $z+w=1+3i$ ; remaining corner $(3,2)$ .

Key idea

Addition preserves a translated arrow's coordinate change, wherever it starts.
Problem 4: Mark complete
Problem 5 A circle record

The figure shows a circle in the complex plane and its center $C$ . Write its equation in the form $|z-c|=r$ . Then decide whether $-1+i$ lies inside, on, or outside the circle.

A circle in the complex plane.

Text description of this figure

A grid with the Real axis from -6 to 2 and the Imaginary axis from -5 to 3, gridlines and number labels at every integer. A circle of radius three grid units is centered at a marked point labeled C, two units left and one unit below the origin. No center coordinates, radius, or equation is printed.
1. Hint 1
  Read the center and measure the radius using equal grid units.
2. Hint 2
  Compare the distance from $-1+i$ to the center with the radius.
Answer

$|z-(-2-i)|=3$ ; $-1+i$ lies inside.

Full solution

The center is $c=-2-i$ and the radius is $3$ , giving
$|z-(-2-i)|=3$
For the tested point, the displacement is $1+2i$ , whose squared length is $5$ .
Since $5<9$ , its distance is less than $3$ , so it lies inside.

Answer

$|z-(-2-i)|=3$ ; $-1+i$ lies inside.

Key idea

A modulus equation specifies all points at a fixed distance from its center.
Problem 5: Mark complete
Problem 6 An unknown scale

Complex numbers $u$ and $v$ satisfy $|u|=\sqrt5$ and $|uv|=15$ . Find $|v|$ exactly, then find $|u\overline v|$ .
1. Hint 1
  The modulus of a product is the product of the moduli.
2. Hint 2
  Conjugation preserves modulus, and $\sqrt5$ is nonzero.
Answer

$|v|=3\sqrt5$ ; $|u\overline v|=15$ .

Full solution

Divide $|uv|=|u||v|$ by the positive number $\sqrt5$ .
$|v|=\frac{15}{\sqrt5}$
Rationalizing gives $3\sqrt5$ .
Since $|\overline v|=|v|$ ,
$|u\overline v|=\sqrt5(3\sqrt5)$
This equals $15$ , checking the product relation.

Answer

$|v|=3\sqrt5$ ; $|u\overline v|=15$ .

Key idea

Conjugation preserves modulus, and modulus is multiplicative.
Problem 6: Mark complete
Problem 7 A shifted circle

Every point $z$ on $|z-(1-2i)|=2$ is moved by adding $-3+i$ . Describe the resulting circle by its center, radius, and a modulus equation in the new point $w$ .
1. Hint 1
  Adding the same number to every point is a rigid slide.
2. Hint 2
  The center moves by the same addition, and distances from it stay fixed.
Answer

Center $-2-i$ , radius $2$ ; $|w-(-2-i)|=2$ .

Full solution

The center moves to
$(1-2i)+(-3+i)=-2-i$
The slide preserves radius $2$ .
Since $w=z-3+i$ , the new displacement from the center equals the old one, so the resulting circle is
$|w-(-2-i)|=2$

Answer

Center $-2-i$ , radius $2$ ; $|w-(-2-i)|=2$ .

Key idea

A translation moves a circle and its center together without changing its radius.
Problem 7: Mark complete
Problem 8 A length equality

For $z=-1+5i$ and $w=3-15i$ , a student claims $|z+w|=|z|+|w|$ because both arrows lie on the same line through zero. Decide whether the claim holds, and explain the directional condition for equality.
1. Hint 1
  Sharing a line does not say whether two arrows point the same way.
2. Hint 2
  Express $w$ as a real multiple of $z$ and inspect the sign.
Answer

False; $|z+w|=2\sqrt{26}$ and $|z|+|w|=4\sqrt{26}$ . Equality needs the same direction, or a zero arrow.

Full solution

Here $w=-3z$ , so the arrows point in opposite directions.
Their sum is $-2z$ , giving
$|z+w|=2\sqrt{26}$
The separate lengths add to
$\sqrt{26}+3\sqrt{26}=4\sqrt{26}.$
The triangle inequality $|z+w|\le |z|+|w|$ becomes equality when the arrows point the same way, or one is zero.

Answer

False; $|z+w|=2\sqrt{26}$ and $|z|+|w|=4\sqrt{26}$ . Equality needs the same direction, or a zero arrow.

Key idea

Triangle-inequality equality depends on direction as well as alignment.
Problem 8: Mark complete
Problem 9 A comparison of markers

A student says $z=4+7i$ and $w=7+4i$ have the same distance from zero but are different numbers. Is the statement correct? Explain, and say whether equal distances would justify writing $z=w$ .
1. Hint 1
  Equality compares parts, while modulus compares distances.
2. Hint 2
  Compare the sums of squares and then compare the coordinate pairs.
Answer

Correct; both moduli are $\sqrt{65}$ , but $z\ne w$ . Equal distances do not justify equality.

Full solution

The two squared moduli are
$\begin{aligned} |z|^2&=4^2+7^2 \\ &=65 \\ |w|^2&=7^2+4^2 \\ &=65 \end{aligned}$
The real parts $4$ and $7$ differ, so the numbers differ even though the moduli match.
Equal modulus locates points on the same circle about zero; it does not identify one point.

Answer

Correct; both moduli are $\sqrt{65}$ , but $z\ne w$ . Equal distances do not justify equality.

Key idea

Equal distance from zero does not imply equal complex coordinates.
Problem 9: Mark complete
Problem 10 Two possible routes

Suppose $|z|=2$ and $|w|=7$ . A report gives $|z+w|=4$ . Is that report possible? Justify your answer by finding both general bounds, and exhibit a pair attaining the lower bound.
1. Hint 1
  The sum length is bounded by the sum of the lengths and by their difference.
2. Hint 2
  The shortest route occurs when the two arrows point in opposite directions.
Answer

Impossible; $5\le |z+w|\le9$ . The lower bound is attained by $z=2$ , $w=-7$ .

Full solution

The triangle inequality gives $|z+w|\le9$ .
Applying it to $w=(z+w)-z$ gives
$7\le |z+w|+2$
Hence $|z+w|\ge5$ , ruling out $4$ .
The pair $z=2$ , $w=-7$ has the required moduli and sum $-5$ , whose modulus is $5$ .

Answer

Impossible; $5\le |z+w|\le9$ . The lower bound is attained by $z=2$ , $w=-7$ .

Key idea

The modulus of a sum lies between the difference and the sum of the two moduli.
Problem 10: Mark complete

← Back to lesson