yda Posted November 17, 2004 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
yda Posted November 17, 2004 Author 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
matt grime Posted November 17, 2004 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.
matt grime Posted November 17, 2004 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.
yda Posted November 17, 2004 Author 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 ?
yda Posted November 17, 2004 Author 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 ?
matt grime Posted November 17, 2004 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.
matt grime Posted November 17, 2004 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.
bloodhound Posted November 17, 2004 Posted November 17, 2004 same here, altough i just plugged the equation in maple and made it solve it.
bloodhound Posted November 17, 2004 Posted November 17, 2004 same here, altough i just plugged the equation in maple and made it solve it.
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