Here's the question you clicked on:
JenniferSmart1
How do I find absolute values of complex numbers? |7-i|
pythatgoras \[|a+bi|=\sqrt{a^2+b^2}\]
in your case \(a=7,b=-1\) so you can just about do it in your head
\[\sqrt{7^2+1^2}=\sqrt{491}=\sqrt{50}=5\sqrt2\]
not much to memorize, the hypotenuse is the square root of the sum of the squares, as in a right triangle
but what has an "i" to do with a triangle?
|dw:1358396646228:dw|
the complex numbers live in the complex plane the absolute value is the distance from the origin, which you get via pythagoras \[a^2+b^2=h^2\] \[h=\sqrt{a^2+b^2}\]
is the triangle always in that quadrant?
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
ahh good to know...so is "a" and "b" the length of the lines?
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 \(7-i\)
|dw:1358397280980:dw|
That makes sense...I just have one more dumb question a+bi....is that the location of the point at the end of that line?
oh no that is the length of the line....durrr..sorry!!!1
the absolute value of a+bi is the length of the line...correct?
a + bi a = number (coordinate) in the real part b = number (coordinate) in the imaginary part | a+bi | = length of the line => yes, I think (Actually... I haven't learnt it in the lesson yet.. So...)
that's cool...It makes sense to me though. Thanks soo much @Callisto =)
yes, it is the length of the line, that is, the distance between the complex number \(a+bi\) and the origin