How do you find the magnitude of a vector with complex components?

Question

Take for instance, a vector v=ax+iy+bz. I'm unsure how to handle the imaginary part. Do you just square it so that the magnitude is Sqrt[a^2-1+b^2] or should i take the absolute value of it's square giving Sqrt[a^2+1+b^2]? Thanks.

Answer 1

I think what you need is a definition for magnitude in this case and I think an appropriate one is sqrt( v dot v_conjugate ). So for your v, it is the second result, but not absolute value. In fact, say (1+i)x + (2+3i)y has magnitude sqrt( 2 + 13) = sqrt(15).



inner product is a key word in this.

Answer 2

Magnitude Of Complex Vector

Answer 3

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Either with complex numbers -2.5m + j 4.2m = 4.89m <121° or magnitude = √ [ (-2.5)^2 + (4.2)^2 ] = 4.89m & angle = arctan (+4.2/-2.5) = -59.2° We have a negative x and a positive y. This puts us in quadrant II, so we have to add 180° to our angle to get the angle relative to the positive x-axis. angle = 180° + -59.2° = 121° Ans: 4.89m <121°

Answer 4

see here how it is defined:

http://cnx.org/content/m10504/latest/