Complex Quadratic Equations Practice Quiz (8 questions)

Question 1

1. For $a=\alpha+i\beta$ and $z=x+iy$ , which system is equivalent to $z^2=a$ ?

x^2-y^2=\alpha

,

x^2+y^2=\alpha^2+\beta^2

, and

2xy=-\beta

x^2+y^2=\alpha

,

x^2-y^2=\sqrt{\alpha^2+\beta^2}

, and

2xy=\beta

x^2-y^2=\alpha

,

x^2+y^2=\sqrt{\alpha^2+\beta^2}

, and

2xy=\beta

x^2+y^2=\alpha

,

x^2-y^2=\beta

, and

2xy=\sqrt{\alpha^2+\beta^2}

Explanation

Expanding $(x+iy)^2$ and matching real and imaginary parts gives the first and third equations, while adding the squared components yields the modulus relation in the first choice. The second choice swaps the roles of the sum and difference, so it does not represent the required system.

Answer

$$x^2-y^2=\alpha$$, $$x^2+y^2=\sqrt{\alpha^2+\beta^2}$$, and $$2xy=\beta$$

Question 2

2. When solving $z^2=\alpha+i\beta$ by writing $z=x+iy$ , which procedure correctly determines the possible square roots?

Add and subtract the component equations to find

x^2

and

y^2

, then choose signs satisfying

2xy=\beta

.

Solve the real equation for

x

, set

y=\beta

, and use the imaginary equation as a check.

Find

x

and

y

from the modulus equation, then assign signs without checking their product.

Equate

x^2+y^2

with the real part and determine the imaginary part from the resulting value of

x-y

.

Explanation

The standard method first obtains the squares of the real and imaginary components by adding and subtracting the equations, then uses the sign condition $2xy=\beta$ . Assigning signs without that condition can produce pairs whose imaginary component has the wrong sign.

Answer

Add and subtract the component equations to find $$x^2$$ and $$y^2$$, then choose signs satisfying $$2xy=\beta$$.

Question 3

3. Which pair gives both solutions of $z^2=3+4i$ ?

2-i

and

-2+i

3+4i

and

-3-4i

2+i

and

-2-i

\sqrt3+2i

and

-\sqrt3-2i

Explanation

Squaring $2+i$ gives $3+4i$ , and the second square root is its opposite, $-2-i$ . The conjugate $2-i$ instead squares to $3-4i$ , so it has the wrong imaginary part.

Answer

$$2+i$$ and $$-2-i$$

Question 4

4. What are the two solutions of $z^2=-6-8i$ ?

-6-8i

and

6+8i

2\sqrt2-\sqrt2i

and

-2\sqrt2+\sqrt2i

\sqrt2+2\sqrt2i

and

-\sqrt2-2\sqrt2i

\sqrt2-2\sqrt2i

and

-\sqrt2+2\sqrt2i

Explanation

For this equation, the component relations give $x^2=2$ and $y^2=8$ , and the negative imaginary part requires $x$ and $y$ to have opposite signs. The third choice uses the same magnitudes but produces a positive imaginary component when the root with positive real part is squared.

Answer

$$\sqrt2-2\sqrt2i$$ and $$-\sqrt2+2\sqrt2i$$

Question 5

5. What is the primary definition of the square root of a complex number $a=\alpha+i\beta$ ?

It is a complex number that, when squared, results in a purely imaginary number.

It is the number that, when multiplied by itself, gives the magnitude of

a

, ignoring the phase or argument.

It is a real number that, when squared, equals

a

, regardless of whether

a

is real or complex.

It is a complex number

z=x+iy

such that

z^2=a

, satisfying the equations

x^2 - y^2 = \alpha

,

x^2 + y^2 = \sqrt{\alpha^2 + \beta^2}

, and

2xy = \beta

.

Explanation

The square root of a complex number $a=\alpha+i\beta$ is defined as a complex number $z=x+iy$ that satisfies $z^2=a$ , which leads to the equations $x^2 - y^2 = \alpha$ , $x^2 + y^2 = \sqrt{\alpha^2 + \beta^2}$ , and $2xy = \beta$ . This set of equations characterizes the solutions for the square roots in the complex plane.

Answer

It is a complex number $$z=x+iy$$ such that $$z^2=a$$, satisfying the equations $$x^2 - y^2 = \alpha$$, $$x^2 + y^2 = \sqrt{\alpha^2 + \beta^2}$$, and $$2xy = \beta$$.

Question 6

6. What is the key equation used to find the square roots of a complex number $a=\alpha+i\beta$ ?

The equations

x^2 + y^2 = \alpha

and

x^2 - y^2 = \sqrt{\alpha^2 + \beta^2}

must be satisfied.

The equations

x^2 - y^2 = \alpha

,

x^2 + y^2 = \sqrt{\alpha^2 + \beta^2}

, and

2xy = \beta

must be satisfied.

The equations

x^2 + y^2 = \alpha^2 + \beta^2

and

x^2 - y^2 = \alpha - \beta

must be satisfied.

The equations

x^2 + y^2 = \alpha

and

x^2 - y^2 = \beta

must be satisfied.

Explanation

The key equations for finding the square roots of a complex number involve expressing the real and imaginary parts as $x^2 - y^2 = \alpha$ and $2xy = \beta$ , along with $x^2 + y^2 = \sqrt{\alpha^2 + \beta^2}$ . The other options do not correctly represent the system derived from equating $z^2$ to $a$ .

Answer

The equations $$x^2 - y^2 = \alpha$$, $$x^2 + y^2 = \sqrt{\alpha^2 + \beta^2}$$, and $$2xy = \beta$$ must be satisfied.

Question 7

7. What is the main purpose of the complex quadratic formula in solving equations of the form $az^2 + bz + c = 0$ ?

To find the roots of the quadratic equation by calculating the square root of the discriminant and applying it to the formula.

To convert the quadratic equation into a linear equation for easier solving.

To identify whether the roots of the quadratic are real or complex without calculating them.

To determine the coefficients

a

,

b

, and

c

from the roots of the quadratic equation.

Explanation

The complex quadratic formula is used to find the roots of the quadratic equation by calculating the square root of the discriminant, which may be complex, and then applying it to the formula. It does not determine coefficients or convert the equation into a linear form, nor does it directly identify the nature of roots without calculation.

Answer

To find the roots of the quadratic equation by calculating the square root of the discriminant and applying it to the formula.

Question 8

8. When was the formula for the roots of a complex quadratic equation, involving the discriminant, first established in mathematical history?

It was first introduced in the 17th century by René Descartes.

The roots formula was formalized in the early 20th century with the advent of modern algebra.

The formula was established in the 19th century during the development of complex analysis.

It was known since ancient Greece, with Euclid's work on quadratic equations.

Explanation

The complex quadratic formula, involving the discriminant, was formalized in the 19th century as part of the development of complex analysis and algebra. Earlier mathematicians like Descartes contributed to quadratic solutions, but the explicit complex form was established later.

Answer

The formula was established in the 19th century during the development of complex analysis.

Complex Quadratic Equations Practice Quiz (8 questions)

All questions and answers

1. For $a=\alpha+i\beta$ and $z=x+iy$ , which system is equivalent to $z^2=a$ ?

2. When solving $z^2=\alpha+i\beta$ by writing $z=x+iy$ , which procedure correctly determines the possible square roots?

3. Which pair gives both solutions of $z^2=3+4i$ ?

4. What are the two solutions of $z^2=-6-8i$ ?

5. What is the primary definition of the square root of a complex number $a=\alpha+i\beta$ ?

6. What is the key equation used to find the square roots of a complex number $a=\alpha+i\beta$ ?

7. What is the main purpose of the complex quadratic formula in solving equations of the form $az^2 + bz + c = 0$ ?

8. When was the formula for the roots of a complex quadratic equation, involving the discriminant, first established in mathematical history?

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Complex Quadratic Equations Practice Quiz (8 questions)

All questions and answers

1. For a=α+iβa=\alpha+i\betaa=α+iβ and z=x+iyz=x+iyz=x+iy, which system is equivalent to z2=az^2=az2=a?

2. When solving z2=α+iβz^2=\alpha+i\betaz2=α+iβ by writing z=x+iyz=x+iyz=x+iy, which procedure correctly determines the possible square roots?

3. Which pair gives both solutions of z2=3+4iz^2=3+4iz2=3+4i?

4. What are the two solutions of z2=−6−8iz^2=-6-8iz2=−6−8i?

5. What is the primary definition of the square root of a complex number a=α+iβa=\alpha+i\betaa=α+iβ?

6. What is the key equation used to find the square roots of a complex number a=α+iβa=\alpha+i\betaa=α+iβ?

7. What is the main purpose of the complex quadratic formula in solving equations of the form az2+bz+c=0az^2 + bz + c = 0az2+bz+c=0?

8. When was the formula for the roots of a complex quadratic equation, involving the discriminant, first established in mathematical history?

Study with flashcards

Read the study guide

Related classes

Agricultural Equipment Evolution

Introduction to Biology

Cell Division: Mitosis and Meiosis

Multivariable Functions and Optimization

Set Theory Fundamentals

The Logic of Life

Make your own practice quizzes

1. For $a=\alpha+i\beta$ and $z=x+iy$ , which system is equivalent to $z^2=a$ ?

2. When solving $z^2=\alpha+i\beta$ by writing $z=x+iy$ , which procedure correctly determines the possible square roots?

3. Which pair gives both solutions of $z^2=3+4i$ ?

4. What are the two solutions of $z^2=-6-8i$ ?

5. What is the primary definition of the square root of a complex number $a=\alpha+i\beta$ ?

6. What is the key equation used to find the square roots of a complex number $a=\alpha+i\beta$ ?

7. What is the main purpose of the complex quadratic formula in solving equations of the form $az^2 + bz + c = 0$ ?