Question 242873
<pre><font size = 4 color = "indigo"><b>
{{{abs(z)=sqrt(x^2+y^2)=3}}}, so {{{x^2+y^2=9}}} and {{{y^2=9-x^2}}}

{{{ z+1/z= (x+iy)+ 1/(x+iy)= (x+iy) + ( 1/(x+iy) ) ((x-iy)/(x-iy)) =
x+iy + (x-iy)/(x^2+y^2)= x+iy + (x-iy)/9= x+iy+x/9-iy/9 = (x+x/9)+(y-y/9)i = (10/9)x+(8/9)yi }}}

Therefore:

{{{abs(z+1/z)=abs((10/9)x+(8/9)yi) = sqrt((100/81)x^2+(64/81)y^2)=(1/9)sqrt(100x^2+64y^2)  =(1/9)sqrt(100x^2+64(9-x^2)) =""}}}

{{{ (1/9)sqrt(100x^2+576-64x^2)=(1/9)sqrt(36x^2+576)= (1/9)sqrt(36(x^2+16))=(6/9)sqrt(x^2+16)=(2/3)sqrt(x^2+16)}}}

This is a minimum when {{{x=0}}}, thus

{{{abs(z+1/z)}}} has minimum value {{{(2/3)(sqrt(0^2+16))=(2/3)sqrt(16)=(2/3)(4)=8/3}}}

Edwin</pre>