Question 175558
The absolute value of any complex number in the form {{{a+bi}}} is {{{abs(a+bi)=sqrt(a^2+b^2)}}}. Remember, absolute value geometrically is the distance from the origin. So in this case, we're finding the distance is from the point (a,b) to the origin (0,0) (in the complex plane)



Example:



Problem:

Find the absolute value of {{{3-6i}}}. 



Solution:


In this case, {{{a=3}}} and {{{b=-6}}}



{{{abs(a+bi)=sqrt(a^2+b^2)}}} Start with the given formula.



{{{abs(3-6i)=sqrt((3)^2+(-6)^2)}}} Plug in {{{a=3}}} and {{{b=-6}}}



{{{abs(3-6i)=sqrt(9+36)}}} Square 3 to get 9. Square -6 to get 36



{{{abs(3-6i)=sqrt(45)}}} Add



{{{abs(3-6i)=3*sqrt(5)}}} Simplify the square root (note: If you need help with simplifying square roots, check out this <a href=http://www.algebra.com/algebra/homework/Radicals/simplifying-square-roots.solver> solver</a>)




So the absolute value of {{{3-6i}}} is {{{3*sqrt(5)}}} which approximates to {{{6.708}}}