Question

How do I find absolute values of complex numbers?
|7-i|

Answer 1

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

Answer 2

*pythagoras

Answer 3

in your case \(a=7,b=-1\) so you can just about do it in your head

Answer 4

\[\sqrt{7^2+1^2}=\sqrt{491}=\sqrt{50}=5\sqrt2\]

Answer 5

not much to memorize, the hypotenuse is the square root of the sum of the squares, as in a right triangle

Answer 6

but what has an "i" to do with a triangle?

Answer 7

the complex numbers live in the complex plane
the absolute value is the distance from the origin, which you get via pythagoras
\[a^2+b^2=h^2\]
\[h=\sqrt{a^2+b^2}\]

Answer 8

is the triangle always in that quadrant?

Answer 9

No, it depends on the complex number.

Let a, b = +ve
a+bi => quad. I
-a + bi => quad. II
-a - bi => quad. III
a - bi => quad. IV

Answer 10

ahh good to know...so is "a" and "b" the length of the lines?

Answer 11

it doesn't make any difference however, what quadrant you are in, it is still
\[|a+bi|=\sqrt{a^2+b^2}\] i just put it there because your number was \(7-i\)