How do you find #abs( 6-3i )#?
2 Answers
Mar 3, 2018
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
Explanation:
Any complex no. of the form
So, here x = 6, y = -3.
So we get: