$(a+ib)^{2002} = (a-ib)$
Multiply by (a-ib) on both sides to get
$(a+ib)^{2003} = (a-ib)(a+ib) = a^2+b^2\cdots(1)$
Now from the given condition we have taking mod on both sides
$(\sqrt{a^2+b^2})^{2002} = \sqrt{a^2+b^2}$
Or $(\sqrt{a^2+b^2})((\sqrt{a^2+b^2}) ^{2002} -1)= 0$
So $(\sqrt{a^2+b^2})= 0\cdots(2)$ or $(\sqrt{a^2+b^2}) = 1\cdots(3)$
From (2) one gives $(a,b) = (0,0)$ one solution
from (1) and (3) we get
$(a+ib)^{2003} = 1$ and this has got 2003 solutions giving total 2004 solutions
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