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Calculus

How to find roots of a complex equation

Alan Smith
Posted 10 years ago

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.
POSTED BY: Alan Smith
3 Replies
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    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 BY: Udo Krause
Alan Smith
Alan Smith
Posted 10 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 BY: Alan Smith 
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.
POSTED BY: Udo Krause
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