MATH 113 Midterm: MATH 113 Harvard 113 Fall 01113Solution2
Document Summary
October 11, 2001: find all complex solutions of ez = 5: ex+iy = ex(cosy + isiny) = 5(cos + isin ). Note: if er1cis 1 = er2cis 2 then r1 = r2 and 1 2 = 2 k for k z. (pf: |ericis i| = eri r1 = r2. Since cos is 1-1 in each quadrant and comparing the signs of sin and cos determine the quadrant of an angle, this implies 1 = 2. ) So z = ln5 + (2n + 1) i: find all complex solutions of sin(z) = 5 eiz + e iz. (eiz)2 10ieiz 1 = 0. Note: the solutions are closed under complex conjugation as ln(5 + 24) = ln(5 . 24) (pf: ln(5 + 24) + ln(5 24) = ln((5 + 24)(5 24) = ln1 = 0. The - de nition of continuity is equivalent to the statement that the inverse image of open sets is open.
