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Anood
04-15-2007, 03:31 PM
How can i find The imaginary part of (1-2i)^2-i.

This is what i have done so far:
(1-2i)^2 (1-2i)^-i

=(1-4i+4)(1-2i)^-i

pka
04-15-2007, 04:07 PM
This is a very difficult problem. There is no easy way to do it.
Here are the basic tools you will need.

\L \begin{array}{l}
{{\rm Im}\nolimits} (zw) = {{\rm Im}\nolimits} (z){{\rm Re}\nolimits} (w) + {{\rm Im}\nolimits} (w){{\rm Re}\nolimits} (z) \\
z^w = e^{w\log (z)} \\
{{\rm Im}\nolimits} \left( {e^z } \right) = e^{{{\rm Re}\nolimits} (z)} \sin (\arg (z)) \\
\left( {1 - 2i} \right)^{\left( { - i} \right)} = e^{ - i\left[ {\log (\sqrt 5 ) + i\left( {\arg (1 - 2i) + 2n\pi } \right)} \right]} \\
\end{array}

BTW \left( {1 - 2i} \right)^2 = - 3 - 4i.

morson
05-13-2007, 08:37 AM
Sorry to revive a dead thread, but I read this thread a while ago and didn't know how to find an expression for (a + bi)^i, but I then read an article about Euler and what he did just recently (he was mind-numbingly brilliant), about his series expansions and expressions for cos(a + bi) and sin(a + bi), and I pieced together this:

http://img519.imageshack.us/img519/4425/eulerbj8.png

But, there's a high chance that this could be wrong.

So, from this:

==> Im[(1 - 2i)^{2-i}] = -5e^{Arg(1-2i)}*sin[Arg(3+4i) - ln{sqrt(5)}]

Now, Arg(1 - 2i) is a negative angle between 0 and -90 degrees, and Arg(3 + 4i) is an acute positive angle.

Of course, if I'm talking nonsense, please let me know.

?

pka
05-13-2007, 09:59 AM
Not sure that I follow what you are trying to do.

The definition given above for the logarithm, w^z = e^{z\log (w)}, is standard. I suggest your working with this to understand some basics.
For example:
i^i = e^{i\log (i)} = e^{i(\ln (1) + i(2k\pi + \pi /2)} = e^{ - (2k\pi + \pi /2)}.
As you can see the imaginary part of that is 0.
The value of the expression for k=0 is known as the principal value.
That corresponds to the Log, logarithm with a capital L.

So here is the principal value of what you posted.
\begin{array}{rcl}
\left( {a + bi} \right)^i & = & e^{iLog(a + ib)} \\
& = & e^{i\left( {\ln \left( {\sqrt {a^2 + b^2 } } \right) + iArg(a + bi)} \right)} \\
& = &e^{ - Arg(a + bi)} \left[ {cis(\left( {\ln \left( {\sqrt {a^2 + b^2 })} \right] \\
\end{array}

Now the imaginary part of the principal value is:
{{\ Im}\nolimits} \left( {\left( {a + bi} \right)^i } \right) = e^{ - Arg(a + bi)} \left[ {\sin (\left( {\ln \left( {\sqrt {a^2 + b^2 })} \right].

EDIT: correction.

morson
05-13-2007, 10:44 AM
\begin{array}{rcl}
\left( {a + bi} \right)^i & = & e^{iLog(a + ib)} \\
& = & e^{i\left( {\ln \left( {\sqrt {a^2 + b^2 } } \right) + iArg(a + bi)} \right)} \\
& = &e^{ - Arg(a + bi)} \left[ {cis(Arg(a + bi))} \right] \\
\end{array}

I am not sure how you got to that last line. I understood everything up to that last part. I presume e^{i*ln(sqrt(a^2+b^2))} to be cis{1/2*ln(a^2+b^2)}, not cis{Arg(a+bi)}.

Would you mind posting your solution for the original problem posed by Anood?

pka
05-13-2007, 11:03 AM
You are correct! I Have edited to correct.
That is what I get for using "copy & paste"!

Would you mind posting your solution for the original problem posed by Anood?Let’s find the principal value.
\begin{array}{rcl}
\left( {1 - 2i} \right)^{2 - i} & = & e^{\left( {2 - i} \right)Log(1 - 2i)} \\
& = & e^{\left( {2 - i} \right)\left[ {\ln (\sqrt 5 ) + i\theta } \right]} \quad ,\quad \theta = \arctan ( - 2) \\
& = & e^{2\ln (\sqrt 5 ) + \theta } \left[ {cis(2\theta - \ln \left( {\sqrt 5 } \right)} \right] \\
\end{array}

\rm Im \left( {e^{2\ln (\sqrt 5 ) + \theta } \left[ {cis(2\theta - \ln \left( {\sqrt 5 } \right)} \right]} \right) = \ln (5)e^\theta \left[ {\sin (2\theta - \ln \left( {\sqrt 5 } \right)} \right]

morson
05-14-2007, 03:47 AM
Okay, I evidently made a mistake somewhere. Thanks.