IB Mathematics: Analysis and Approaches Quiz: Complex Numbers

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Complex NumbersQuestion 1 of 20

The locus of a complex number $z$ is given by $\arg(z-2) - \arg(z+2i) = \frac{\pi}{2}$ . Which of the following describes this locus?

A straight line

A full circle

A major arc of a circle

A minor arc of a circle

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IB Mathematics: Analysis and Approaches Quiz

IB Mathematics: Analysis and Approaches Quiz: Complex Numbers

Practice Complex Numbers in IB Mathematics: Analysis and Approaches with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Complex Numbers, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Mathematics: Analysis and Approaches.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

The locus of a complex number $z$ is given by $\arg(z-2) - \arg(z+2i) = \frac{\pi}{2}$ . Which of the following describes this locus?

A straight line
A full circle
A major arc of a circle (correct answer)
A minor arc of a circle

Explanation: The given equation can be rewritten using the property of arguments:

\arg\left(\frac{z-2}{z+2i}\right) = \frac{\pi}{2}

.\nLet

w = \frac{z-2}{z+2i}

. The condition

\arg(w) = \frac{\pi}{2}

means that

w

is a purely imaginary number with a positive imaginary part. Let

w=ki

for some

k>0

.\nLet

z = x+iy

. Then

w = \frac{(x-2)+iy}{x+i(y+2)}

.\nWe set the real part of

w

to zero. The real part of a quotient

\frac{a+bi}{c+di}

\frac{ac+bd}{c^2+d^2}

. For this to be zero, we only need

ac+bd=0

.\nHere,

a=x-2, b=y, c=x, d=y+2

.\nSo,

(x-2)x + y(y+2) = 0

.\n

x^2-2x + y^2+2y = 0

.\nCompleting the square gives:

(x^2-2x+1) + (y^2+2y+1) - 1 - 1 = 0

, which is

(x-1)^2+(y+1)^2=2

.\nThis is the equation of a circle with centre

(1,-1)

and radius

\sqrt{2}

.\nHowever, we also have the condition that

w

has a positive imaginary part. The imaginary part of the quotient is

\frac{bc-ad}{c^2+d^2}

. We need

bc-ad>0

.\n

yx - (x-2)(y+2) > 0

xy - (xy+2x-2y-4) > 0

-2x+2y+4 > 0 \implies y-x+2 > 0 \implies y > x-2

.\nThis inequality restricts the solution to a portion of the circle. The locus starts at

z=-2i

and ends at

z=2

(these points are excluded). The circle passes through these two points. The arc that satisfies

y>x-2

is the major arc of the circle connecting these two points.

Question 2

The complex numbers $z_1 = 3+2i$ and $z_2 = -1+4i$ represent two adjacent vertices of a square in the complex plane. Which of the following complex numbers could represent another vertex of the square?

-3 (correct answer)
1-2i
2+6i
5+6i

Explanation: Let the vertices be A(

\small z_1

) and B(

\small z_2

). The vector from A to B is represented by the complex number

v = z_2 - z_1 = (-1+4i) - (3+2i) = -4+2i

. \nThe other two vertices, C and D, form the square ABCD. The vector BC must be perpendicular to AB and have the same length. We can find the vector BC by rotating the vector BA by

\pm 90^{\circ}

(

\pm \frac{\pi}{2}

) about B. The vector BA is

-v = 4-2i

.\nRotation by

90^{\circ}

counter-clockwise corresponds to multiplication by

i

. Rotation by

90^{\circ}

clockwise corresponds to multiplication by

-i

.\n\textbf{Case 1:} Rotate BA by

90^{\circ}

counter-clockwise.\nVector BC

= i \times (z_1-z_2) = i(4-2i) = 4i - 2i^2 = 2+4i

.\nThe position of vertex C is

z_3 = z_2 + (2+4i) = (-1+4i) + (2+4i) = 1+8i

.\n\textbf{Case 2:} Rotate BA by

90^{\circ}

clockwise.\nVector BC

= -i \times (z_1-z_2) = -i(4-2i) = -4i + 2i^2 = -2-4i

.\nThe position of vertex C is

z_3 = z_2 + (-2-4i) = (-1+4i) + (-2-4i) = -3

.\nSo, two possible locations for a third vertex are

1+8i

and

-3

. The fourth vertex would be

z_4 = z_1 + \vec{BC}

. In Case 1,

z_4 = (3+2i)+(2+4i) = 5+6i

. In Case 2,

z_4 = (3+2i)+(-2-4i) = 1-2i

. The possible vertices are

1+8i

5+6i

-3

, and

1-2i

. From the options,

-3

is a possible vertex.

Question 3

-3 (correct answer)
1-2i
2+6i
5+6i

Explanation: Let the vertices be A(

\small z_1

) and B(

\small z_2

). The vector from A to B is represented by the complex number

v = z_2 - z_1 = (-1+4i) - (3+2i) = -4+2i

. \nThe other two vertices, C and D, form the square ABCD. The vector BC must be perpendicular to AB and have the same length. We can find the vector BC by rotating the vector BA by

\pm 90^{\circ}

(

\pm \frac{\pi}{2}

) about B. The vector BA is

-v = 4-2i

.\nRotation by

90^{\circ}

counter-clockwise corresponds to multiplication by

i

. Rotation by

90^{\circ}

clockwise corresponds to multiplication by

-i

.\n\textbf{Case 1:} Rotate BA by

90^{\circ}

counter-clockwise.\nVector BC

= i \times (z_1-z_2) = i(4-2i) = 4i - 2i^2 = 2+4i

.\nThe position of vertex C is

z_3 = z_2 + (2+4i) = (-1+4i) + (2+4i) = 1+8i

.\n\textbf{Case 2:} Rotate BA by

90^{\circ}

clockwise.\nVector BC

= -i \times (z_1-z_2) = -i(4-2i) = -4i + 2i^2 = -2-4i

.\nThe position of vertex C is

z_3 = z_2 + (-2-4i) = (-1+4i) + (-2-4i) = -3

.\nSo, two possible locations for a third vertex are

1+8i

and

-3

. The fourth vertex would be

z_4 = z_1 + \vec{BC}

. In Case 1,

z_4 = (3+2i)+(2+4i) = 5+6i

. In Case 2,

z_4 = (3+2i)+(-2-4i) = 1-2i

. The possible vertices are

1+8i

5+6i

-3

, and

1-2i

. From the options,

-3

is a possible vertex.

Question 4

Find the general form of all complex numbers $z$ that satisfy the equation $z^2 + |z|^2 = 0$ .

$z = ki$ for any $k \in \mathbb{R}$ (correct answer)
$z=0$ only
$z = k$ for any $k \in \mathbb{R}$
$z = k(1+i)$ for any $k \in \mathbb{R}$

Explanation: When you encounter equations involving both complex numbers and their modulus (absolute value), you need to work with both the algebraic and geometric properties of complex numbers. Let's write

z = a + bi

where

a, b \in \mathbb{R}

. Then

|z|^2 = a^2 + b^2

, so our equation becomes:

(a + bi)^2 + a^2 + b^2 = 0

Expanding the left side:

a^2 + 2abi + (bi)^2 + a^2 + b^2 = 0

a^2 + 2abi - b^2 + a^2 + b^2 = 0

2a^2 + 2abi = 0

2a(a + bi) = 0

This gives us

a = 0

a + bi = 0

. If

a + bi = 0

, then

z = 0

. If

a = 0

, then

z = bi

for any real

b

, which we can write as

z = ki

where

k \in \mathbb{R}

. Let's verify: if

z = ki

, then

z^2 = -k^2

and

|z|^2 = k^2

, so

z^2 + |z|^2 = -k^2 + k^2 = 0

✓ Choice A is correct - it captures all purely imaginary numbers (including

z = 0

when

k = 0

). Choice B is wrong because it only includes zero, missing all other purely imaginary solutions. Choice C is wrong because real numbers

z = k

would give

k^2 + k^2 = 2k^2 = 0

, which only works for

k = 0

. Choice D is wrong because

z = k(1+i)

gives

2k^2i + 2k^2 \neq 0

for

k \neq 0

. Study tip: When solving complex equations with modulus, separate real and imaginary parts systematically - this reveals the geometric constraint that solutions must lie on specific lines or curves in the complex plane.

Question 5

The transformation $T$ of the complex plane is defined by $w = (1+i)z + 2$ . Which sequence of transformations maps a point $z$ to $w$ ?

A translation by $\begin{pmatrix} 2 \\ 0 \end{pmatrix}$ , then a rotation by $\frac{\pi}{4}$ counter-clockwise about the origin and an enlargement of scale factor $\sqrt{2}$ from the origin.
A rotation by $\frac{\pi}{4}$ counter-clockwise about the origin, an enlargement of scale factor $\sqrt{2}$ from the origin, followed by a translation by $\begin{pmatrix} 2 \\ 0 \end{pmatrix}$ . (correct answer)
A rotation by $\frac{\pi}{2}$ counter-clockwise about the origin, an enlargement of scale factor $2$ from the origin, followed by a translation by $\begin{pmatrix} 2 \\ 0 \end{pmatrix}$ .
A rotation by $\frac{\pi}{4}$ counter-clockwise about the origin, an enlargement of scale factor $\sqrt{2}$ from the origin, followed by a translation by $\begin{pmatrix} 0 \\ 2 \end{pmatrix}$ .

Explanation: The transformation is

w = (1+i)z + 2

. This is of the form

w = az+b

, which represents a rotation and dilation, followed by a translation. The order of operations is multiplication first, then addition.\n1. Multiplication by $a=1+i$ : We write

1+i

in polar form.

|1+i| = \sqrt{1^2+1^2} = \sqrt{2}

and

\arg(1+i) = \frac{\pi}{4}

. So,

1+i = \sqrt{2}e^{i\pi/4}

. Multiplication by this complex number corresponds to an enlargement (dilation) by a scale factor of

\sqrt{2}

from the origin and a rotation by

\frac{\pi}{4}

counter-clockwise about the origin. These two can be performed in any order.\n2. Addition of $b=2$ : Adding 2 corresponds to a translation by the vector

\begin{pmatrix} 2 \\ 0 \end{pmatrix}

.\nSo the full transformation is a rotation and enlargement, followed by a translation. Choice B correctly describes this sequence. Choice A has the wrong order: translating first would lead to

w = (1+i)(z+2) = (1+i)z + 2+2i

, which is a different transformation.

Question 6

Let the complex number $z$ be given by $z = \frac{2+ai}{1-2i}$ , where $a \in \mathbb{R}$ . If the real part of $z$ is zero, find the value of $a$ .

-4
-2
1 (correct answer)
4

Explanation: To find the real part of

z

, we first write

z

in the form

x+yi

by multiplying the numerator and denominator by the conjugate of the denominator, which is

1+2i

.\n

z = \frac{2+ai}{1-2i} \times \frac{1+2i}{1+2i} = \frac{(2+ai)(1+2i)}{1^2 + (-2)^2} = \frac{2+4i+ai+2ai^2}{1+4} = \frac{(2-2a)+(4+a)i}{5}

\nSo,

z = \frac{2-2a}{5} + i\frac{4+a}{5}

.\nThe real part of

z

\text{Re}(z) = \frac{2-2a}{5}

.\nWe are given that

\text{Re}(z) = 0

, so we set the real part equal to zero and solve for

a

:\n

\frac{2-2a}{5} = 0 \implies 2-2a = 0 \implies 2 = 2a \implies a=1

Question 7

Find the sum of the infinite geometric series $1 + \frac{1+i}{2} + (\frac{1+i}{2})^2 + (\frac{1+i}{2})^3 + \dots$ .

i
1-i
1+i (correct answer)
The series diverges.

Explanation: This is an infinite geometric series with first term

a=1

and common ratio

r = \frac{1+i}{2}

.\nFor the series to converge, the modulus of the common ratio must be less than 1 (

|r|<1

).\n

|r| = |\frac{1+i}{2}| = \frac{|1+i|}{2} = \frac{\sqrt{1^2+1^2}}{2} = \frac{\sqrt{2}}{2}

.\nSince

\sqrt{2} \approx 1.414

|r| \approx 0.707 < 1

, so the series converges.\nThe sum to infinity of a convergent geometric series is given by the formula

S = \frac{a}{1-r}

.\n

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

\nTo express this in the form

x+yi

, we multiply the numerator and denominator by the conjugate of the denominator:\n

S = \frac{2}{1-i} \times \frac{1+i}{1+i} = \frac{2(1+i)}{1^2 - i^2} = \frac{2(1+i)}{1 - (-1)} = \frac{2(1+i)}{2} = 1+i

Question 8

Let $z_1 = 1+i\sqrt{3}$ and $z_2 = 1-i$ . Find the principal argument of $\frac{z_1^4}{z_2^2}$ .

-\frac{5\pi}{6}
-\frac{\pi}{6} (correct answer)
\frac{5\pi}{6}
\frac{11\pi}{6}

Explanation: We use the property

\arg(\frac{w_1}{w_2}) = \arg(w_1) - \arg(w_2)

and

\arg(z^n) = n\arg(z)

.\nFirst, find the arguments of

z_1

and

z_2

.\nFor

z_1 = 1+i\sqrt{3}

, it is in the first quadrant.

\arg(z_1) = \arctan(\frac{\sqrt{3}}{1}) = \frac{\pi}{3}

.\nFor

z_2 = 1-i

, it is in the fourth quadrant.

\arg(z_2) = \arctan(\frac{-1}{1}) = -\frac{\pi}{4}

.\nNow, we find the argument of the expression:\n

\arg(\frac{z_1^4}{z_2^2}) = \arg(z_1^4) - \arg(z_2^2) = 4\arg(z_1) - 2\arg(z_2)

= 4(\frac{\pi}{3}) - 2(-\frac{\pi}{4}) = \frac{4\pi}{3} + \frac{\pi}{2} = \frac{8\pi}{6} + \frac{3\pi}{6} = \frac{11\pi}{6}

.\nThis is a valid argument, but it is not the principal argument, which must be in the interval

(-\pi, \pi]

. To find the principal argument, we add or subtract multiples of

2\pi

.\n

\frac{11\pi}{6} - 2\pi = \frac{11\pi}{6} - \frac{12\pi}{6} = -\frac{\pi}{6}

.\nSince

-\pi < -\frac{\pi}{6} \le \pi

, the principal argument is

-\frac{\pi}{6}

Question 9

Let $\omega$ be a non-real cube root of unity. Find the value of $(1-\omega+\omega^2)(1+\omega-\omega^2)$ .

-4
0
3
4 (correct answer)

Explanation: The cube roots of unity are the solutions to

z^3=1

, or

z^3-1=0

. This factors as

(z-1)(z^2+z+1)=0

. The roots are

z=1

and the roots of

z^2+z+1=0

. Since

\omega

is a non-real root, it must satisfy

\omega^2+\omega+1=0

. From this identity, we can derive two useful relations:\n1)

1+\omega^2 = -\omega

\n2)

1+\omega = -\omega^2

\nNow we substitute these into the expression we want to evaluate:\n

(1-\omega+\omega^2)(1+\omega-\omega^2)

\nFor the first factor:

(1+\omega^2) - \omega = (-\omega) - \omega = -2\omega

.\nFor the second factor:

(1+\omega) - \omega^2 = (-\omega^2) - \omega^2 = -2\omega^2

.\nSo the product is

(-2\omega)(-2\omega^2) = 4\omega^3

.\nSince

\omega

is a cube root of unity, we know that

\omega^3=1

.\nTherefore, the value of the expression is

4(1)=4

Question 10

A complex number $z$ has modulus 4 and argument $\frac{2\pi}{3}$ . Find $z - 2\bar{z}$ in the form $a+bi$ .

-6 - 2i\sqrt{3}
-2 + 6i\sqrt{3}
2 - 2i\sqrt{3}
2 + 6i\sqrt{3} (correct answer)

Explanation: First, we express

z

in Cartesian form

a+bi

.\nGiven

|z|=4

and

\arg(z) = \frac{2\pi}{3}

.\n

z = |z|(\cos(\arg(z)) + i\sin(\arg(z)))

z = 4(\cos(\frac{2\pi}{3}) + i\sin(\frac{2\pi}{3}))

\nWe know that

\cos(\frac{2\pi}{3}) = -\frac{1}{2}

and

\sin(\frac{2\pi}{3}) = \frac{\sqrt{3}}{2}

.\nSo,

z = 4(-\frac{1}{2} + i\frac{\sqrt{3}}{2}) = -2 + 2i\sqrt{3}

.\nNext, we find the conjugate of

z

\bar{z}

.\n

\bar{z} = -2 - 2i\sqrt{3}

.\nNow, we calculate the expression

z - 2\bar{z}

:\n

z - 2\bar{z} = (-2 + 2i\sqrt{3}) - 2(-2 - 2i\sqrt{3})

= -2 + 2i\sqrt{3} + 4 + 4i\sqrt{3}

\nCombine the real and imaginary parts:\n

= (-2+4) + (2\sqrt{3}+4\sqrt{3})i

= 2 + 6i\sqrt{3}

Question 11

A polynomial $P(z) = z^3 + az^2 + bz + 15$ has real coefficients $a$ and $b$ . Given that $z_1 = 2+i$ is a root of $P(z)$ , find the value of $a$ .

-5
-1 (correct answer)
1
5

Explanation: Since the polynomial

P(z)

has real coefficients, if

z_1 = 2+i

is a root, then its complex conjugate

z_2 = 2-i

must also be a root.\nFor a cubic polynomial, there must be a third root,

z_3

. According to Vieta's formulas, the product of the roots is

z_1z_2z_3 = -\frac{15}{1} = -15

.\nFirst, we find the product of the known roots:

z_1z_2 = (2+i)(2-i) = 2^2 - i^2 = 4 - (-1) = 5

.\nNow we can find the third root:

5 \cdot z_3 = -15 \implies z_3 = -3

.\nAlso from Vieta's formulas, the sum of the roots is

z_1+z_2+z_3 = -\frac{a}{1} = -a

.\nSumming the three roots:

(2+i) + (2-i) + (-3) = 4 - 3 = 1

.\nTherefore,

-a = 1

, which means

a = -1

Question 12

A complex number $z$ satisfies the equation $|z - 3i| = 2|z|$ . The locus of points representing $z$ in the complex plane is a circle. Find the centre of this circle.

-2i
-i (correct answer)
i
2i

Explanation: Let

z = x+iy

, where

x, y \in \mathbb{R}

. Substitute this into the given equation:\n

|(x+iy) - 3i| = 2|x+iy|

|x + i(y-3)| = 2|x+iy|

\nNow, apply the definition of the modulus:

|a+bi| = \sqrt{a^2+b^2}

.\n

\sqrt{x^2 + (y-3)^2} = 2\sqrt{x^2+y^2}

\nSquare both sides to eliminate the square roots:\n

x^2 + (y-3)^2 = 4(x^2+y^2)

x^2 + y^2 - 6y + 9 = 4x^2 + 4y^2

\nRearrange the terms to get the equation of a circle:\n

3x^2 + 3y^2 + 6y - 9 = 0

\nDivide the entire equation by 3:\n

x^2 + y^2 + 2y - 3 = 0

\nTo find the centre, we complete the square for the

y

terms:\n

x^2 + (y^2 + 2y + 1) - 1 - 3 = 0

x^2 + (y+1)^2 = 4

\nThis is the equation of a circle with centre

(0, -1)

and radius 2. In the complex plane, the centre

(0, -1)

corresponds to the complex number

0 - 1i = -i

Question 13

How many distinct non-zero complex numbers $z$ satisfy the equation $z^2 = \bar{z}$ ?

1
2
3 (correct answer)
4

Explanation: Let's solve the equation

z^2 = \bar{z}

using polar form. Let

z = re^{i\theta}

. Then

\bar{z} = re^{-i\theta}

.\nThe equation becomes

(re^{i\theta})^2 = re^{-i\theta}

, which simplifies to

r^2e^{i2\theta} = re^{-i\theta}

.\nWe can equate the moduli and the arguments.\nEquating moduli:

r^2 = r \implies r^2 - r = 0 \implies r(r-1) = 0

. This gives

r=0

r=1

.\nIf

r=0

, then

z=0

. This is one solution.\nIf

r=1

, we equate the arguments:

e^{i2\theta} = e^{-i\theta}

. This implies that

2\theta = -\theta + 2k\pi

for some integer

k

.\n

3\theta = 2k\pi \implies \theta = \frac{2k\pi}{3}

.\nWe find the distinct values for

\theta

by taking

k=0, 1, 2

.\nFor

k=0

\theta=0

. This gives the solution

z = 1e^{i0} = 1

.\nFor

k=1

\theta=\frac{2\pi}{3}

. This gives the solution

z = e^{i2\pi/3} = -\frac{1}{2} + i\frac{\sqrt{3}}{2}

.\nFor

k=2

\theta=\frac{4\pi}{3}

. This gives the solution

z = e^{i4\pi/3} = -\frac{1}{2} - i\frac{\sqrt{3}}{2}

.\nFor

k=3

, we get

\theta = 2\pi

, which is the same as

\theta=0

.\nThe solutions are

0, 1, e^{i2\pi/3}, e^{i4\pi/3}

. The question asks for the number of non-zero solutions. There are 3 non-zero solutions.

Question 14

How many distinct solutions does the equation $z|z| - 2z + i = 0$ have?

0
1
2 (correct answer)
3

Explanation: Rearrange the equation to isolate

z

:\n

z(|z|-2) = -i

.\nFrom this form, we can see that

z = \frac{-i}{|z|-2}

. Since the right side is a real multiple of

-i

z

must be a purely imaginary number. Let

z=iy

for some real number

y

.\nThe modulus is

|z| = |iy| = |y|

.\nSubstitute

z=iy

and

|z|=|y|

into the original equation:\n

iy|y| - 2(iy) + i = 0

\nSince

i \neq 0

, we can divide the entire equation by

i

:\n

y|y| - 2y + 1 = 0

.\nWe must consider two cases based on the sign of

y

.\nCase 1: $y \ge 0$ \nIn this case,

|y|=y

. The equation becomes:\n

y(y) - 2y + 1 = 0 \implies y^2 - 2y + 1 = 0

(y-1)^2 = 0 \implies y=1

.\nSince

y=1

satisfies the condition

y \ge 0

, this gives a valid solution

z = i(1) = i

.\nCase 2: $y < 0$ \nIn this case,

|y|=-y

. The equation becomes:\n

y(-y) - 2y + 1 = 0 \implies -y^2 - 2y + 1 = 0

y^2 + 2y - 1 = 0

.\nUsing the quadratic formula to solve for

y

:\n

y = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2(1)} = \frac{-2 \pm \sqrt{8}}{2} = \frac{-2 \pm 2\sqrt{2}}{2} = -1 \pm \sqrt{2}

.\nWe have two potential values for

y

y_1 = -1+\sqrt{2} \approx 0.414

and

y_2 = -1-\sqrt{2} \approx -2.414

.\nSince this case requires

y<0

, we must discard

y_1

. The only valid solution in this case is

y_2 = -1-\sqrt{2}

, which gives the solution

z = i(-1-\sqrt{2})

.\nIn total, we have two distinct solutions:

z=i

and

z = (-1-\sqrt{2})i

Question 15

Find the value of the sum $S = \sum_{n=1}^{101} i^n$ .

-1
0
1
i (correct answer)

Explanation: The sum is

S = i + i^2 + i^3 + i^4 + \dots + i^{101}

. The powers of

i

form a cycle of length 4:

i, -1, -i, 1

. The sum of each cycle is

i - 1 - i + 1 = 0

.\nWe can group the terms in sets of four. The number of terms is 101.\n

101 = 4 \times 25 + 1

.\nThis means we have 25 full cycles of four terms, plus one extra term.\nThe sum of the first 100 terms is

\sum_{n=1}^{100} i^n = 25 \times (i+i^2+i^3+i^4) = 25 \times 0 = 0

.\nSo, the sum

S

is just the last term,

i^{101}

.\nTo evaluate

i^{101}

, we use the remainder when 101 is divided by 4.\n

101 = 4 \times 25 + 1

, so

i^{101} = (i^4)^{25} \cdot i^1 = 1^{25} \cdot i = i

.\nTherefore,

S = i

Question 16

A polynomial $P(z) = z^3 + az^2 + bz + 15$ has real coefficients $a$ and $b$ . Given that $z_1 = 2+i$ is a root of $P(z)$ , find the value of $a$ .

-5
-1 (correct answer)
1
5

Explanation: Since the polynomial

P(z)

has real coefficients, if

z_1 = 2+i

is a root, then its complex conjugate

z_2 = 2-i

must also be a root.\nFor a cubic polynomial, there must be a third root,

z_3

. According to Vieta's formulas, the product of the roots is

z_1z_2z_3 = -\frac{15}{1} = -15

.\nFirst, we find the product of the known roots:

z_1z_2 = (2+i)(2-i) = 2^2 - i^2 = 4 - (-1) = 5

.\nNow we can find the third root:

5 \cdot z_3 = -15 \implies z_3 = -3

.\nAlso from Vieta's formulas, the sum of the roots is

z_1+z_2+z_3 = -\frac{a}{1} = -a

.\nSumming the three roots:

(2+i) + (2-i) + (-3) = 4 - 3 = 1

.\nTherefore,

-a = 1

, which means

a = -1

Question 17

Find the area of the polygon whose vertices in the complex plane are given by the roots of the equation $z^4 = -16$ .

4
8 (correct answer)
16
32

Explanation: First, we find the roots of

z^4 = -16

. We express -16 in polar form. The modulus is 16 and the argument is

\pi

. So,

-16 = 16e^{i\pi}

.\nLet

z = re^{i\theta}

. Then

z^4 = r^4e^{i4\theta}

.\nComparing with

16e^{i(\pi+2k\pi)}

, we have:\n

r^4 = 16 \implies r=2

.\n

4\theta = \pi + 2k\pi \implies \theta = \frac{\pi+2k\pi}{4}

for

k=0, 1, 2, 3

.\nThe arguments are:

\theta_0 = \frac{\pi}{4}

\theta_1 = \frac{3\pi}{4}

\theta_2 = \frac{5\pi}{4}

\theta_3 = \frac{7\pi}{4}

.\nThe four roots all lie on a circle of radius 2 centred at the origin. They are equally spaced angularly, with a separation of

\frac{2\pi}{4} = \frac{\pi}{2}

. Therefore, the vertices form a square inscribed in a circle of radius 2.\nThe polygon can be seen as four isosceles triangles with the origin as a common vertex. Each triangle has two sides of length 2 (the radius) and the angle between them is

\frac{\pi}{2}

. The area of one such triangle is

A_{triangle} = \frac{1}{2}ab\sin C = \frac{1}{2}(2)(2)\sin(\frac{\pi}{2}) = 2(1) = 2

.\nSince there are four such triangles, the total area of the square is

4 \times 2 = 8

Question 18

Find the sum of the infinite geometric series $1 + \frac{1+i}{2} + (\frac{1+i}{2})^2 + (\frac{1+i}{2})^3 + \dots$ .

i
1-i
1+i (correct answer)
The series diverges.

Explanation: This is an infinite geometric series with first term

a=1

and common ratio

r = \frac{1+i}{2}

.\nFor the series to converge, the modulus of the common ratio must be less than 1 (

|r|<1

).\n

|r| = |\frac{1+i}{2}| = \frac{|1+i|}{2} = \frac{\sqrt{1^2+1^2}}{2} = \frac{\sqrt{2}}{2}

.\nSince

\sqrt{2} \approx 1.414

|r| \approx 0.707 < 1

, so the series converges.\nThe sum to infinity of a convergent geometric series is given by the formula

S = \frac{a}{1-r}

.\n

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

\nTo express this in the form

x+yi

, we multiply the numerator and denominator by the conjugate of the denominator:\n

S = \frac{2}{1-i} \times \frac{1+i}{1+i} = \frac{2(1+i)}{1^2 - i^2} = \frac{2(1+i)}{1 - (-1)} = \frac{2(1+i)}{2} = 1+i

Question 19

The equation $z^4 - 2z^3 + 7z^2 - 4z + 10 = 0$ has a root $z_1 = 1-2i$ . Find a root of the equation that is purely imaginary.

2i
i\sqrt{2} (correct answer)
i\sqrt{5}
2i\sqrt{2}

Explanation: The polynomial has real coefficients, so if

z_1 = 1-2i

is a root, its conjugate

z_2 = 1+2i

must also be a root.\nThis means that

(z - (1-2i))(z - (1+2i))

is a factor of the polynomial.\nThis factor is

z^2 - (z_1+z_2)z + z_1z_2

.\nSum of roots:

z_1+z_2 = (1-2i)+(1+2i) = 2

.\nProduct of roots:

z_1z_2 = (1-2i)(1+2i) = 1^2 - (2i)^2 = 1 - (-4) = 5

.\nSo,

z^2 - 2z + 5

is a factor.\nWe can find the other factor by polynomial long division:\n

(z^4 - 2z^3 + 7z^2 - 4z + 10) \div (z^2 - 2z + 5)

.\nThe division gives

z^2+2

.\nSo,

z^4 - 2z^3 + 7z^2 - 4z + 10 = (z^2 - 2z + 5)(z^2+2)

.\nThe remaining roots are the solutions to

z^2+2=0

.\n

z^2 = -2 \implies z = \pm\sqrt{-2} = \pm i\sqrt{2}

.\nThese two roots,

i\sqrt{2}

and

-i\sqrt{2}

, are purely imaginary. Therefore, one such root is

i\sqrt{2}

Question 20

A complex number $z$ satisfies the equation $2z + i\bar{z} = 5-i$ . Find $z$ .

3 - i
3 + i
\frac{11}{3} - \frac{7}{3}i (correct answer)
\frac{11}{3} + \frac{7}{3}i

Explanation: Let

z = x+iy

, where

x, y \in \mathbb{R}

. Then its conjugate is

\bar{z} = x-iy

.\nSubstitute these into the equation:\n

2(x+iy) + i(x-iy) = 5-i

\nDistribute the terms:\n

2x + 2iy + ix - i^2y = 5-i

\nSince

i^2 = -1

, this becomes:\n

2x + 2iy + ix + y = 5-i

\nGroup the real and imaginary parts on the left side:\n

(2x+y) + i(x+2y) = 5-i

\nNow, equate the real and imaginary parts from both sides of the equation:\nReal part:

2x+y = 5

(1)\nImaginary part:

x+2y = -1

(2)\nWe have a system of two linear equations. From (1), we can write

y = 5-2x

. Substitute this into (2):\n

x + 2(5-2x) = -1

x + 10 - 4x = -1

-3x = -11 \implies x = \frac{11}{3}

\nNow substitute the value of

x

back into the expression for

y

:\n

y = 5 - 2(\frac{11}{3}) = 5 - \frac{22}{3} = \frac{15-22}{3} = -\frac{7}{3}

\nSo,

z = x+iy = \frac{11}{3} - \frac{7}{3}i