How do you find #abs( -4+2i )#?

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Answer 1 · 2016-05-26T04:11:21

#|-4+2i|=2sqrt5~=4.5#

We have the complex number

#c=-4+2i#

There are two equivalent expressions for the magnitude of an imaginary number, one in terms of the real and imaginary parts and

#|c|=+sqrt{RRe(c)^2 + Im(c)^2}# ,

and another in terms of the complex conjugate

# # # # # # # # #=+ sqrt(c*bar{c})# .

I'm going to use the first expression because it's simpler, in certian cases the 2nd may be more useful.

We need the real part and imaginary parts of #-4+2i#
#RRe(-4+2i)=-4#
#Im(-4+2i)=2#
#|-4+2i|=sqrt{(-4)^2+(2)^2}=sqrt{16+4}=sqrt{20}=2sqrt5~=4.5#