yda Posted November 17, 2004 Share Posted November 17, 2004 Im trying to solve this equation for along time I tries to use the formulas for power 3 but it ddint help I think it shuld be factoring the left side but I couldnt make it z^9-z'^9=9|z|^4 please help Link to comment Share on other sites More sharing options...
yda Posted November 17, 2004 Author Share Posted November 17, 2004 Im trying to solve this equation for along time I tries to use the formulas for power 3 but it ddint help I think it shuld be factoring the left side but I couldnt make it z^9-z'^9=9|z|^4 please help Link to comment Share on other sites More sharing options...
matt grime Posted November 17, 2004 Share Posted November 17, 2004 What does that dash mean? (z'^9) Presumably complex conjugate. LHS is 2*Im(z^9) which, if z is re^(i\theta), is 2r^9sin(\theta), and that equals 9r^4. Link to comment Share on other sites More sharing options...
matt grime Posted November 17, 2004 Share Posted November 17, 2004 What does that dash mean? (z'^9) Presumably complex conjugate. LHS is 2*Im(z^9) which, if z is re^(i\theta), is 2r^9sin(\theta), and that equals 9r^4. Link to comment Share on other sites More sharing options...
yda Posted November 17, 2004 Author Share Posted November 17, 2004 I tried it in the trigonometric way : 2r^9[isin(9*theta)]=9|z|^4 but where should I go from here ? Link to comment Share on other sites More sharing options...
yda Posted November 17, 2004 Author Share Posted November 17, 2004 I tried it in the trigonometric way : 2r^9[isin(9*theta)]=9|z|^4 but where should I go from here ? Link to comment Share on other sites More sharing options...
matt grime Posted November 17, 2004 Share Posted November 17, 2004 yep, correction to me, the LHS isn't what I sad. Note that the LHS is purely imaginary and the RHS purely real, thus there are no solutions except the trivial one, z=0, assuming z' means conjugate. Link to comment Share on other sites More sharing options...
matt grime Posted November 17, 2004 Share Posted November 17, 2004 yep, correction to me, the LHS isn't what I sad. Note that the LHS is purely imaginary and the RHS purely real, thus there are no solutions except the trivial one, z=0, assuming z' means conjugate. Link to comment Share on other sites More sharing options...
yda Posted November 17, 2004 Author Share Posted November 17, 2004 You are right thanks Link to comment Share on other sites More sharing options...
yda Posted November 17, 2004 Author Share Posted November 17, 2004 You are right thanks Link to comment Share on other sites More sharing options...
bloodhound Posted November 17, 2004 Share Posted November 17, 2004 same here, altough i just plugged the equation in maple and made it solve it. Link to comment Share on other sites More sharing options...
bloodhound Posted November 17, 2004 Share Posted November 17, 2004 same here, altough i just plugged the equation in maple and made it solve it. Link to comment Share on other sites More sharing options...
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