Question 945908
This includes finding the size of a complex number.  Maybe easier to do that first.


{{{abs(2-3i)=sqrt(2^2+3^2)=sqrt(4+9)=sqrt(13)}}}


Revise original expression:


{{{(1/2)sqrt(sqrt(13)+a)+- (i*sqrt(sqrt(13))-a)}}}


{{{sqrt(sqrt(13)+a)/2+- (i*sqrt(sqrt(13)-a)}}}


{{{sqrt(sqrt(13)+a)/2+i*sqrt(sqrt(13))-a}}}
OR
{{{sqrt(sqrt(13)+a)/2-i*sqrt(sqrt(13))+a}}}


Putting into better complex number form,
{{{(sqrt(sqrt(13)+a-2a)/2+i*sqrt(sqrt(13))}}}
OR
{{{(sqrt(sqrt(13)+a+2a)/2-i*sqrt(sqrt(13))}}}


Finally:
{{{highlight((sqrt(sqrt(13)-a)/2+i*sqrt(sqrt(13)))}}}
OR
{{{highlight((sqrt(sqrt(13)+3a)/2-i*sqrt(sqrt(13)))}}}


(Look at the text under the failed-renduring to see clearly.)