# How to find roots of a complex equation

Posted 8 years ago
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 I'm trying to find the roots of sin z, (= sin(x+iy) I tried things like 'solve [ sin(x+iy)=0,x,y]' - but I get an error - "(1-y) is not a valid variable" - I agree :-) So 2 things please, 1st correct my expression above, 2nd are there better ways of doing this? Thanks.
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Posted 8 years ago
  In[4]:= Reduce[Sin[x + I y] == 0, {x, y}] Out[4]= C[1] \[Element] Integers && (y == I x - 2 I \[Pi] C[1] || y == I x - I (\[Pi] + 2 \[Pi] C[1])) In[7]:= FindInstance[Sin[x + I y] == 0, {x, y}, Complexes] Out[7]= {{x -> 0, y -> 0}} using Mathematica 10.1.0.0. Just for the logs: lower case reduce and lower case sin are no built-in commands. If you defined them they possibly need a fix. If you use special initialization codes, try to Switch them off.Another Thing to try is: Edit -> Preferences -> Reset to Defaults.
Posted 8 years ago
 Hi & thanks - this sounds real good, but when I try what you have - "reduce [ sin[x+Iy]==0, {x,y}] ", I still get the error '1-y is not a valid variable. '? I think I have cleared all variables etc., so not sure what needs fixing my end?I also tried "FindInstance [ sin[x+Iy]==0, {x,y}, Complexes]" - but got a wrong answer - {{y -> ConditionalExpression[ 1/2 (1 - 2 \[Pi] C[1] + I (1 - 2 \[Pi] C[1])), C[1] \[Element] Integers]}, {y -> ConditionalExpression[ 1/2 (1 - \[Pi] - 2 \[Pi] C[1] + I (1 - \[Pi] - 2 \[Pi] C[1])), C[1] \[Element] Integers]}} It should be x=n[pi], y=0.....? Thanks
Posted 8 years ago
 In[2]:= Reduce[Sin[x + I y] == 0, {x, y}] Out[2]= C[1] \[Element] Integers && (y == I x - 2 I \[Pi] C[1] || y == I x - I (\[Pi] + 2 \[Pi] C[1])) Solve heads to algebraic equation (Systems), Reduce to transcendent ones. See also FindInstance, please.