How do you find #abs( 6-3i )#?

2 Answers
Mar 3, 2018

#3sqrt5#

Explanation:

#"given a complex number "z=x+yi" then"#

#•color(white)(x)|z|=|x+yi|=sqrt(x^2+y^2)#

#"here "x=6" and "y=-3#

#rArr|6-3i|=sqrt(36+9)=sqrt45=3sqrt5#

Mar 3, 2018

#3sqrt 5#

Explanation:

Any complex no. of the form #z = x+iy# has #|z| =sqrt( x^2 + y^2)#
So, here x = 6, y = -3.
So we get:
#|z| = sqrt(6^2 + (-3)^2) = sqrt(36 + 9) = sqrt(45) = 3sqrt5#