Question 375712
{{{(2w+i)/(w-i) = (3w+4i)/(w+3i)}}}.  Cross-multiply to get
{{{(2w+i)(w+3i)= (3w+4i)(w-i) }}}, 
{{{2w^2 + 6iw + iw + 3i^2 = 3w^2 - 3iw + 4iw - 4i^2}}}, or
{{{2w^2 +7iw - 3 = 3w^2 + iw + 4}}}, or
{{{0 = w^2 - 6iw + 7}}},
0 = (w -7i )(w + i), giving
w = 7i or w = -i.  Since none of these values make the denominators of the original equation equal to zero, these are the final answers.