# Complex Numbers - Absolute Values

The absolute value of a number is often viewed as the "distance" a number is away from 0, the origin.

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## General Concepts

For real numbers, the absolute value is just the magnitude of the number without considering its sign. For example, the absolute value of -5 is 5, and the absolute value of 5 is also 5.

For a complex number $z = a + bi$ represented on the complex plane by the pair $(a, b)$, the "distance" from the origin is found using the Pythagorean theorem. The absolute value of $z$ is defined as

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

For example, the absolute value of the complex number $3+4i$ is equal to

$|3+4i| = \sqrt{3^2 +4^2} = 5.$

The absolute value can also be written as

$|z| = \sqrt{z \bar{z}},$

where$\bar{z}$ is the complex conjugate of $z.$

## Examples/Problems

What is the absolute value of the complex number $-5+12i?$

We have $\lvert -5+12i \rvert=\sqrt{(-5)^2+12^2}=\sqrt{25+144}=\sqrt{169}=\sqrt{13^2}=13.$ $_\square$

The absolute value of the complex number $7+bi$ is $\sqrt{170}.$ What is the negative number $b?$

The absolute value of $7+bi$ is $\lvert 7+bi \rvert =\sqrt{7^2+b^2}=\sqrt{170},$ implying $7^2+b^2=170,$ or $b^2=121.$ This implies $b=\pm 11.$ Since $b$ is negative, $b=-11.$ $_\square$

The absolute value of the complex number $-7-8i$ can be expressed as $\lvert -7-8i \rvert=\sqrt{(-7-8i)(a+bi)},$ where $a$ and $b$ are real numbers and $i$ is the imaginary number. What are $a$ and $b?$

Observe that the absolute value of a complex number $z$ can be written as $\lvert z \rvert = \sqrt{z \bar{z}},$ where$\bar{z}$ is the complex conjugate of $z.$ Then since $z=-7-8i,$ thus $a+bi=-7+8i,$ which implies that $a=-7$ and $b=8.$ $_\square$

## What is the absolute value of the following sum of complex numbers: $(12+4i)-(9-13i)-3i?$

We have $\lvert (12+4i)-(9-13i)-3i \rvert = \lvert 3+14i \rvert =\sqrt{3^2+14^2}=\sqrt{205}.$ $_\square$

## Let $(6, 8)$ be the coordinates of the complex number $z$ on the complex plane. Then what is the absolute value of its complex conjugate $\bar{z}?$

Since $z=6+8i,$ it follows that $\bar{z}=6-8i.$ Then $\lvert \bar{z} \rvert=\lvert6-8i \rvert=\sqrt{(6)^2+(-8)^2}=\sqrt{36+64}=\sqrt{100}=\sqrt{10^2}=10. \ _\square$

## The complex number $a+bi$ has the same absolute value as a different complex number $3+4i.$ If $a$ and $b$ are both positive integers, what are $a$ and $b?$

The absolute value of $3+4i$ is $\lvert 3+4i \rvert =\sqrt{3^2+4^2}=\sqrt{25}=5.$ Since $a+bi$ also has an absolute value of $5,$ it follows that $\lvert a+bi \rvert =\sqrt{a^2+b^2}=5,$ implying $a^2+b^2=25.$ Since $a$ and $b$ are both positive integers and $a+bi\ne 3+4i$ by assumption, the only way that $a^2+b^2=25$ holds is that $a=4$ and $b=3.$ $_\square$

## See Also

**Cite as:**Complex Numbers - Absolute Values.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/complex-numbers-absolute-values/