Complex Numbers

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Problems about complex numbers.

A rational number is any number that can be expressed as $\answer {\frac {a}{b}}$ where $a$ and $b$ are $\answer [format=string]{integers}$ and $b$ $=$ $<$ $>$ $\ne$ $\answer {0}$ .

The $\answer [format=string]{real}$ numbers are all of the numbers on the number line.

Correct! This is why the number line is sometimes called the real line.

Real numbers that are not rational are said to be $\answer [format=string]{irrational}$ .

Complex numbers can be expressed as $a+bi$ , where $a$ and $b$ are $\answer [format=string]{real}$ numbers.

Which of the following are rational numbers? Select all that apply.

$7$ $e$ $\frac {\pi ^2}{6}$ $\frac {18}{11}$ $8-3i$ $\sqrt {-17}$ $\sqrt [3]{-2}$

Which of the following are real numbers? Select all that apply.

$7$ $e$ $\frac {\pi ^2}{6}$ $\frac {18}{11}$ $8-3i$ $\sqrt {-17}$ $\sqrt [3]{-2}$

Which of the following are complex numbers? Select all that apply.

$7$ $e$ $\frac {\pi ^2}{6}$ $\frac {18}{11}$ $8-3i$ $\sqrt {-17}$ $\sqrt [3]{-2}$

Complex numbers can be expressed as $a+bi$ , where $a$ and $b$ are real numbers.

Correct! The complex numbers include the real numbers.

A complex number that is not real is often called $\answer [format=string]{imaginary}$ .

Suppose $z=a+bi$ , where $a$ and $b$ are real numbers. The number $z$ is called real when $\answer {b}=0$ or pure imaginary when $\answer {a}=0$ .

Assuming none of the numbers involved are zero, select all operations below that must produce an irrational number.

rational $+$ rational rational $+$ irrational irrational $+$ irrational rational $\times$ rational irrational $\times$ rational irrational $\times$ irrational

In the following problems express the result in the form $a+bi$ , with $a$ and $b$ real numbers.

Compute $(2+i) + 4 =$ $\answer [given]{6} + \answer [given]{1}\,i$

Compute $(-3+4i) - (-8 - i) =$ $\answer [given]{5} + \answer [given]{5}\,i$

Compute $(2-6i) - (3+8i) =$ $\answer [given]{-1} + \answer [given]{-14}\,i$

Compute $(2+i) \cdot 4 =$ $\answer [given]{8} + \answer [given]{4}\,i$

Compute $(-3 + 4i) \cdot (-8 - i) =$ $\answer [given]{28} + \answer [given]{-29}\,i$

Compute $(2-6i) \cdot (3+8i)=$ $\answer [given]{54} + \answer [given]{-2}\,i$

Compute $(2-3i) \cdot (2-3i)=$ $\answer [given]{-5} + \answer [given]{-12}\,i$

Compute $(2-3i) \cdot (2+3i)=$ $\answer [given]{13} + \answer [given]{0}\,i$

Compute $(a+bi) \cdot (a+bi)=$ $\left (\answer [given]{a^2-b^2}\right ) + \left (\answer [given]{2ab}\right )i$

Compute $(a+bi) \cdot (a-bi)=$ $\left (\answer [given]{a^2+b^2}\right ) + \left (\answer [given]{0}\right )i$

Usually the product of two imaginary numbers is an imaginary number. But here we have multiplied the number $a+bi$ by its conjugate, $a-bi$ , and the result is a $\answer [format=string]{real}$ number because the imaginary part of the product is $\answer {0}$ .

Compute $\frac {1}{2+i}=$ $\answer [given]{\frac {2}{5}} + \answer [given]{-\frac {1}{5}}\,i$

Try multiplying the numerator and denominator by something that will make the denominator into a real number.

Try the complex conjugate of $2+i$ .

Compute $(1-3i) \div (-3+5i)=$ $\answer [given]{-\frac {18}{34}} + \answer [given]{\frac {4}{34}}\,i$

Perhaps first compute $\frac {1}{-3+5i}$ , and then multiply.

From your experience with imaginary roots of polynomials, you might guess that if $3+5i$ is a root of the polynomial, then $\answer {3-5i}$ is also a root.

Correct. More generally, if a polynomial has $\answer [format=string]{real}$ coefficients and $a+bi$ is a root, then $\answer {a-bi}$ is also a root. In such cases, we say that the imaginary roots come as a conjugate pair, $a\pm bi$ .

Find all solutions to the equation $x^3-3x^2+5x-3=0$ .

The Rational Root Theorem combined with some division of polynomials might help!

What kinds of roots did you find?

3 real solutions 4 real solutions 1 real solution and 2 imaginary solutions 2 (distinct) real solutions and 2 imaginary solutions 1 (double) real solution and 2 imaginary solutions

Enter first the real solution then the imaginary solutions (as a conjugate pair).

$\answer [given]{1}$ , $\answer [given]{1} \pm \answer [given]{\sqrt {2}}\,i$

Find all solutions to the equation $x^3+2x-3=0$ .

The Rational Root Theorem combined with some division of polynomials might help!

What kinds of roots did you find?

3 real solutions 4 real solutions 1 real solution and 2 imaginary solutions 2 (distinct) real solutions and 2 imaginary solutions 1 (double) real solution and 2 imaginary solutions

Enter first the real solution then the imaginary solutions (as a conjugate pair).

$\answer [given]{1}$ , $\answer [given]{-\frac {1}{2}} \pm \answer [given]{\frac {\sqrt {11}}{2}}\,i$