Question 737751
<pre>
The other tutor's answer is wrong.

Apparently, you're an electronics student who uses i for current, and 
therefore has to use j for &#8730;<span style="text-decoration: overline">-1</span>.

Sorry, I thought you wanted j<sup>j</sup> and worked that out.  I'll redo it
and put the results here later.  It's actually 4.810477381.  But
here is {{{j^j}}}

{{{j^j}}}{{{""=""}}}{{{

matrix(2,1,"",e^ln(j^j)  )




}}}{{{""=""}}}{{{e^(j*ln(j))}}}

Now to continue we must find an expression for ln(j)

Euler's equation is 

{{{e^(j*theta)}}}{{{""=""}}}{{{cos(theta)+j*sin(theta)}}}

Substitute {{{theta=pi/2}}}

{{{matrix(2,1,"",e^(j*expr(pi/2)))}}}{{{""=""}}}{{{cos(pi/2)+j*sin(pi/2)}}}{{{""=""}}}{{{0+j=j}}}

Therefore since 

{{{matrix(2,1,"",e^(j*expr(pi/2)))}}}{{{""=""}}}{{{j}}}

then {{{ln(j)}}}{{{""=""}}}{{{j*expr(pi/2)}}}

Now we go back to where we were:
{{{j^j}}}{{{""=""}}}{{{matrix(2,1,"",e^(ln(j^j)))}}}{{{""=""}}}{{{e^( j*ln(j) )}}}{{{""=""}}}{{{matrix(2,1,"",e^( j*j*expr(pi/2)) )}}}{{{""=""}}}{{{matrix(2,1,"",e^(-1*expr(pi/2)))}}}{{{""=""}}}{{{matrix(2,1,"",e^(-pi/2))}}} = 0.2078795764

Edwin</pre>