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matsorz
Aug27-11, 06:57 AM
I have a couple of problems with the conjugate. I have two equations to solve, (conjugate)z=2/z and (conjugate)z=-2/z
How do I solve these= Are there some rules when you use the conjugate?
tiny-tim
Aug27-11, 07:11 AM
hi matsorz! :smile:
z + \bar{z} = 2Re(z)
z - \bar{z} = 2Im(z)
z\bar{z} = |z|^2 :wink:
matsorz
Aug27-11, 07:26 AM
Ok, so I put |z^2|=2 in a) and |z^2|=-2 in b? Do I put in x and y, or do i just solve for z straight away?
HallsofIvy
Aug27-11, 08:52 AM
One problem you have is that the second equation is impossible.
\overline{z}= -\frac{2}{z}
is the same as z\overline{z}= -2 but z\overline{z}= |z|^2 must be a positive real number.
stallionx
Aug27-11, 09:20 AM
I have a couple of problems with the conjugate. I have two equations to solve, (conjugate)z=2/z and (conjugate)z=-2/z
How do I solve these= Are there some rules when you use the conjugate?
(a-bi)*(a+bi)=2
a**2 + b**2 = 2
center ( 0,0 ) radius =sqrt2
(a-bi)*(a+bi)=-2
center (a,b) radius = i sqrt2
take abs of radii both
one is sqrt2
other abs ((sqrt 2) *i)=abs(i) * abs(sqrt2)
abs(i)=1
so they turn out to be equal.
tiny-tim
Aug27-11, 10:39 AM
Ok, so I put |z^2|=2 in a) and |z^2|=-2 in b? Do I put in x and y, or do i just solve for z straight away?
straight away …
(it's always quickest to avoid coordinates as far as you can)
|z| = √2 (the first case), so the general solution for z is … ? :smile:
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