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Exercise [04.03]
http://www.roadtoreality.info/viewtopic.php?f=19&t=1442
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Author:  JoshSmith [ 25 Nov 2009, 08:10 ]
Post subject:  Exercise [04.03]

Hopefully this helps some people (like me) rusty with their algebra.

Show a+ib=\left(\sqrt{\frac{1}{2}(a+\sqrt{a^2+b^2})}+i\sqrt{\frac{1}{2}(-a+\sqrt{a^2+b^2})}\right)^2.

This requires a fair amount of algebraic expansion:

a+ib=\Bigg(\sqrt{\frac{1}{2}(a+\sqrt{a^2+b^2})}+i\sqrt{\frac{1}{2}(-a+\sqrt{a^2+b^2})}\Bigg)^2
=\Bigg(\sqrt{\frac{1}{2}(a+\sqrt{a^2+b^2})}+i\sqrt{\frac{1}{2}(-a+\sqrt{a^2+b^2})}\Bigg)\cdot\Bigg(\sqrt{\frac{1}{2}(a+\sqrt{a^2+b^2})}+i\sqrt{\frac{1}{2}(-a+\sqrt{a^2+b^2})}\Bigg)
=\frac{1}{2}(a+\sqrt{a^2+b^2})+2\left(i\sqrt{\frac{1}{2}(-a+\sqrt{a^2+b^2}})\sqrt{\frac{1}{2}(a+\sqrt{a^2+b^2})}\right)+i^2\frac{1}{2}(-a+\sqrt{a^2+b2})
=\frac{1}{2}(a+\sqrt{a^2+b^2})+2i\left(\sqrt{(-\frac{a}{2}+\frac{\sqrt{a^2+b^2}}{2})(\frac{a}{2}+\frac{\sqrt{a^2+b^2}}{2})\right)-\frac{1}{2}(-a+\sqrt{a^2+b2})
=\left(\frac{a}{2}+\frac{\sqrt{a^2+b^2}}{2}+\frac{a}{2}-\frac{\sqrt{a^2+b2}}{2}\right)+2i\left(\sqrt{(-\frac{a}{2}+\frac{\sqrt{a^2+b^2}}{2})(\frac{a}{2}+\frac{\sqrt{a^2+b^2}}{2})\right)
=a+2i\left(\sqrt{(-\frac{a}{2}+\frac{\sqrt{a^2+b^2}}{2})(\frac{a}{2}+\frac{\sqrt{a^2+b^2}}{2})\right)
=a+2i\sqrt{-\frac{a^2}{4}+\frac{a^2+b^2}{4}}
=a+2i\sqrt{\frac{b^2}{4}}
=a+2i\frac{b}{2}
=a+ib~,

as we intended to show.

I've intentionally left out some intermediary steps. If you can't work through these intermediaries, I highly suggest doing some more work in algebra. I.M. Gelfand's text has served me well.

Author:  vasco [ 30 Nov 2009, 10:44 ]
Post subject:  Re: Exercise [04.03]

Hi Josh
To help people follow the algebra, it might be useful to point out that you used the identity:

(u+v)^2=u^2+v^2+2uv

and then later:

(u+v)(u-v)=u^2-v^2

Also, I think it makes the algebra much simpler, if, right from the start, you let:

q=\sqrt{(a^2+b^2)}

which gives:

\left(\sqrt{\frac{1}{2}(q+a)}+i\sqrt{\frac{1}{2}(q-a)}\right)^2

and then do the algebra using the above identities, and substitute for q at the end.

Author:  Langing [ 12 Dec 2010, 22:33 ]
Post subject:  Re: Exercise [04.03]

Thanks to vasco for showing the benefit of simplifying before going through the algebra.

Note, there is a second part to this exercise, namely to show that the NEGATIVE of the given complex number also squares to a + ib.

Using vasco's simplification, I was able to quickly verify this for myself, starting by assuming that the NEGATIVE of the given complex number would have minus signs where there are plus signs for the real and imaginary parts.

I also tested the complex conjugate of the given complex number (plus, minus) to see if it might also square to a + ib. Its square turned out to be a - ib, which no doubt has some significance.

Author:  vasco [ 13 Dec 2010, 07:52 ]
Post subject:  Re: Exercise [04.03]

As Latex isn't working let's write cc(p) for complex conjugate of p and x means multiplication


Yes, what you have shown there is that if w=z^2 then cc(w)=cc(z)^2.
This can be shown more directly as follows, if you assume that cc(u x v)=cc(u) x cc(v), which is easy to prove:
w=z^2 so
cc(w)=cc(z^2)=cc(z x z)=cc(z) x cc(z)=cc(z)^2

If you think of it geometrically it is easy to see. if z=a x exp(itheta) then squaring z gives
a^2 x exp(2itheta). So the squaring operation rotates the vector z through theta in an anticlockwise sense (so angle is doubled). If you consider the complex conjugates then these are the same 2 vectors reflected in the x-axis. Draw a picture and you'll see it straight away.

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