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imaginary number equation

In physics, I have come across two equations for impedance:

z=R+jXz =R + j X where j=1j =\sqrt{-1}

z=(R2+X2)z =\sqrt{(R^2 + X^2)}

Does this mean that from:

c2=a2+b2c^2 = a^2 + b^2

then

(a2+b2)a+ib\sqrt{(a^2 + b^2)}\equiv a + ib ??

so is this an alternative form of pythag or am I completely wrong??
Reply 1
Original post by Random999
In physics, I have come across two equations for impedance:

z=R+jXz =R + j X where j=1j =\sqrt{-1}

z=(R2+X2)z =\sqrt{(R^2 + X^2)}

Does this mean that from:

c2=a2+b2c^2 = a^2 + b^2

then

(a2+b2)a+ib\sqrt{(a^2 + b^2)}\equiv a + ib ??

so is this an alternative form of pythag or am I completely wrong??


No, I think you've omitted some modulus signs around z like this: |z|
If z=x+iy (where x and y are both real), then |z|2=x2+y2. As davros says, you appear to have lost a modulus sign somewhere: in general, it is not the case that sqrt(a2+b2)=a+bi.
Reply 3
Original post by Random999

(a2+b2)a+ib\sqrt{(a^2 + b^2)}\equiv a + ib ??

As above, though it may be worth noting that
In general:
Where
Unparseable latex formula:

z = a+bi \ \ a,b \in \matbb{R}

then:
 zz=(a+bi)(abi)=a2+b2\ zz^* = (a +bi)(a-bi) = a^2+b^2

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