Question 62915
YES!!!
:
{{{25/(1+2i)}}}
{{{25(1-2i)/((1+2i)(1-2i))}}}
{{{(25-50i)/(1-2i+2i-4i^2)}}}
{{{(25-50i)/(1-4i^2)}}}  i^2=-1
{{{(25-50i)/(1-4(-1))}}}
{{{(25-50i)/(1+4)}}}
{{{(25-50i)/5}}}
{{{25/5-50i/5}}}
{{{5-10i}}}
Happy Calculating!!!