Quiz: Complex Quadratic Equations — 8 questions

Detailed questions and answers

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

x2y2=αx^2-y^2=\alpha, x2+y2=α2+β2x^2+y^2=\alpha^2+\beta^2, and 2xy=β2xy=-\beta
x2+y2=αx^2+y^2=\alpha, x2y2=α2+β2x^2-y^2=\sqrt{\alpha^2+\beta^2}, and 2xy=β2xy=\beta
x2y2=αx^2-y^2=\alpha, x2+y2=α2+β2x^2+y^2=\sqrt{\alpha^2+\beta^2}, and 2xy=β2xy=\beta
x2+y2=αx^2+y^2=\alpha, x2y2=βx^2-y^2=\beta, and 2xy=α2+β22xy=\sqrt{\alpha^2+\beta^2}

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

Explanation

Expanding (x+iy)2(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.

2. What is the primary definition of the square root of a complex number a=α+iβa=\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 aa, ignoring the phase or argument.
It is a real number that, when squared, equals aa, regardless of whether aa is real or complex.
It is a complex number z=x+iyz=x+iy such that z2=az^2=a, satisfying the equations x2y2=αx^2 - y^2 = \alpha, x2+y2=α2+β2x^2 + y^2 = \sqrt{\alpha^2 + \beta^2}, and 2xy=β2xy = \beta.

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=α+iβa=\alpha+i\beta is defined as a complex number z=x+iyz=x+iy that satisfies z2=az^2=a, which leads to the equations x2y2=αx^2 - y^2 = \alpha, x2+y2=α2+β2x^2 + y^2 = \sqrt{\alpha^2 + \beta^2}, and 2xy=β2xy = \beta. This set of equations characterizes the solutions for the square roots in the complex plane.

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

Add and subtract the component equations to find x2x^2 and y2y^2, then choose signs satisfying 2xy=β2xy=\beta.
Solve the real equation for xx, set y=βy=\beta, and use the imaginary equation as a check.
Find xx and yy from the modulus equation, then assign signs without checking their product.
Equate x2+y2x^2+y^2 with the real part and determine the imaginary part from the resulting value of xyx-y.

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

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=β2xy=\beta. Assigning signs without that condition can produce pairs whose imaginary component has the wrong sign.

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

The equations x2+y2=αx^2 + y^2 = \alpha and x2y2=α2+β2x^2 - y^2 = \sqrt{\alpha^2 + \beta^2} must be satisfied.
The equations x2y2=αx^2 - y^2 = \alpha, x2+y2=α2+β2x^2 + y^2 = \sqrt{\alpha^2 + \beta^2}, and 2xy=β2xy = \beta must be satisfied.
The equations x2+y2=α2+β2x^2 + y^2 = \alpha^2 + \beta^2 and x2y2=αβx^2 - y^2 = \alpha - \beta must be satisfied.
The equations x2+y2=αx^2 + y^2 = \alpha and x2y2=βx^2 - y^2 = \beta 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.

Explanation

The key equations for finding the square roots of a complex number involve expressing the real and imaginary parts as x2y2=αx^2 - y^2 = \alpha and 2xy=β2xy = \beta, along with x2+y2=α2+β2x^2 + y^2 = \sqrt{\alpha^2 + \beta^2}. The other options do not correctly represent the system derived from equating z2z^2 to aa.

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

2i2-i and 2+i-2+i
3+4i3+4i and 34i-3-4i
2+i2+i and 2i-2-i
3+2i\sqrt3+2i and 32i-\sqrt3-2i

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

Explanation

Squaring 2+i2+i gives 3+4i3+4i, and the second square root is its opposite, 2i-2-i. The conjugate 2i2-i instead squares to 34i3-4i, so it has the wrong imaginary part.

6. What is the main purpose of the complex quadratic formula in solving equations of the form az2+bz+c=0az^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 aa, bb, and cc from the roots of the quadratic equation.

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

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.

7. What are the two solutions of z2=68iz^2=-6-8i?

68i-6-8i and 6+8i6+8i
222i2\sqrt2-\sqrt2i and 22+2i-2\sqrt2+\sqrt2i
2+22i\sqrt2+2\sqrt2i and 222i-\sqrt2-2\sqrt2i
222i\sqrt2-2\sqrt2i and 2+22i-\sqrt2+2\sqrt2i

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

Explanation

For this equation, the component relations give x2=2x^2=2 and y2=8y^2=8, and the negative imaginary part requires xx and yy 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.

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.

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

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.

Review with flashcards

Memorize the answers with 11 flashcards on Complex Quadratic Equations.

What equations correspond to the square root of a complex number a=α+iβa=\alpha+i\beta?

They are x2y2=αx^2 - y^2 = \alpha, x2+y2=α2+β2x^2 + y^2 = \sqrt{\alpha^2 + \beta^2}, and 2xy=β2xy = \beta.

Square root of complex number: a=α+iβ

Solve for z=x+iy: equations for x, y from complex parts.

What is the first step to solve z2=az^2 = a for complex zz?

Write zz as x+iyx + iy and equate real and imaginary parts.

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