Problem:
There is a complex number z with imaginary part 164 and a positive integer n such that
z+nz=4i
Find n.
Solution:
Let z=a+bi. Then z=a+bi=(z+n)4i=−4b+4i(a+n). Thus a=−4b and b=4(a+n)=4(n−4b). Solving the last equation for n yields n=4b+4b=4164+4⋅164, so n=697.
The problems on this page are the property of the MAA's American Mathematics Competitions