Question 1115020
{{{z*i=(a+bi)*i}}}
{{{z*i=a*ii+b*i^2}}}
{{{z*i=-b+ai}}}
For this to be real and positive, 
{{{b<0}}}
.
.
.
{{{z*(3-i)=(a+bi)*(3-ii)}}}
{{{z*(3-i)=3a-ai+3bi-bi^2}}}
{{{z*(3-i)=(3a+b)+(3b-a)i}}}
For this to be imaginary,
{{{3a+b=0}}}
and
{{{3b-a<>0}}}