Divide imaginary number

2 + 6i/5 + 2i

Type answer in form a + bi

Do you mean:
(2 + 6i) / (5 + 2i)
= (2 + 6i)(5 - 2i) / (5 + 2i)(5 - 2i)
= (10 - 4i + 30i - 12i^2) / (25 - 10i + 10i - 4i^2)
= (10 + 26i + 12) / (25 + 4)
= (22 + 26i) / 29
= (22/29) + (26/29)i

it "cannot be done" but the trick is to multiply by ( complex conjugate) / (CC)

CC of a + ib = ( a - bi)
the denominator becomes "real" = a^2 + b^2 and the Numerator is complex

Complex numbers are a strange form of vectors

(2+6i)/(5+2i) x (5-2i)/(5-2i) = (22 + 26i)/29 = (10+30i-4i-12i^2)/(5^2 - 4i^2) = (22/29) + (26/29)i.