ketz
  • ketz
explain why j2 = -1 and not 1?
Mathematics
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schrodinger
  • schrodinger
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anonymous
  • anonymous
i believe the definition comes from the complex variable plane, where Im is the y axis and x is Re, then magnitude of a unit vector along -1 on Im axis is defined as sqrt(-1) I'm not sure if that's correct though, look up complex plane
anonymous
  • anonymous
Is it *i* in place of *j*??? @ketz ... It should be... 'i' is the positive square root of -1...
ketz
  • ketz
yes j instead of i...

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anonymous
  • anonymous
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whpalmer4
  • whpalmer4
Electrical engineers typically use \(j^2 = -1\) instead of \(i^2 = -1\)
anonymous
  • anonymous
It's just the definition.
anonymous
  • anonymous
That's how it is defined. j^2 = -1
anonymous
  • anonymous
a complex number is represented as \[z=x+iy\] where x is the real part and y is the imaginary part. all real numbers R are a subset of complex numbers C, for example number 3 has x=3 and y=0 so that only the real part is left. the definition of i (or j) = sqrt(-1) comes from multiplication rule for complex numbers. \[z_1*z_2=(x_1,y_1)(x_2,y_2)=(x_1x_2 - y_1y_2,x_1y_2+x_2y_1)\] if z1=(0,1) and z2=(0,1), that is they both only contain imaginary parts, then \[i^2=(0,1)(0,1)=(-1,0)=-1\] as you can see that after multiplying them out following the formula above we are left with a real part that equals to -1
anonymous
  • anonymous
as a side note, the reason electrical engineers use j instead of i is because they use i to represent current

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