Here two numbers are involved

One is #5# which is easy to write in trigonometric form as

#5=5(1+i0)=5(cos0^@+isin0^@)# or #5e^(i0)#

For #2+3i#, we know #sqrt(2^2+3^2)=sqrt(4+9)=sqrt13#

Hence #2+3i=sqrt13(2/sqrt13+i3/sqrt13)#

Now let #sinalpha=3/sqrt13# and #cosalpha=2/sqrt13#, then

#2+3i=sqrt13(cosalpha+isinalpha)=5e^(ialpha)#

and #5/(2+3i)=(5e^(i0))/(sqrt13e^(ialpha)#

= #5/sqrt13xxe^(0-ialpha)=5/sqrt13e^(-ialpha)#

= #5/sqrt13(cos(-alpha)+isin(-alpha))#

= #5/sqrt13(cosalpha-isinalpha)#

= #5/sqrt13(2/sqrt13-ixx3/sqrt13)#

= #5/13(2-3i)=10/13-i15/13#