71For the first part of problem, I would like to ask whether you tried the direct substitution of z = x+ iy ?
If not , then please try and tell me. If it doesn't works I'll try to think more.
1708\hspace{-16}$Here $\mathbf{z^3+iz=1}$\\\\ Put $\mathbf{z=x+iy}\;,$ Where $\mathbf{x=\mathbb{R}e(z)}$ and $\mathbf{y=\mathbb{I}m(z)}$\\\\ Now Let We assume that The Given eqn. is Intersect $\mathbf{\mathbb{X}-}$axis\\\\ Then Put $\mathbf{y=0}$\\\\ So Put $\mathbf{z=x+iy=x}$ in $\mathbf{z^3+iz=1}$\\\\ $\mathbf{x^3+ix=1\Leftrightarrow ix=(1-x^3)}$\\\\ Now Here $\mathbf{L.H.S}$ is Imaginary and $\mathbf{R.H.S}$ is Real\\\\ So no point of Intersection on $\mathbf{\mathbb{X}-}$axis\\\\ In a Similar way no pint of Intersection on $\mathbf{\mathbb{Y}-}$axis
1708
man111 singh
·2012-01-06 06:00:09
\hspace{-16} $For Calculation of $\mathbf{\mid z \mid}$\\\\ Put $\mathbf{z=x+iy}$ in $\mathbf{z^3+iz=1}$\\\\ $\mathbf{(x+iy)^3+i.(x+iy)=1}$\\\\ $\mathbf{x^3-iy^3+3ixy(x+iy)+ix-y=1}$\\\\ $\mathbf{(x^3-3xy^2)-i.(y^2-3x^2y)=(1+y)-ix}$\\\\ Now Equating Real and Imaginary part, We Get\\\\ \begin{Bmatrix} \bold{x^3-3xy^2=1+y...................(1)\times y} \\\\ \bold{y^3-3x^2y=x.........................(2)\times x} \end{Bmatrix}\\\\\\ Add $\mathbf{(1)+(2)}$\\\\ $\mathbf{-2xy.(x^2+y^2)=y+(y^2+x^2)}$\\\\ $\mathbf{(x^2+y^2).(1+2xy)=-y}$\\\\ $\mathbf{x^2+y^2=\frac{-y}{1+2xy}}$\\\\\\ $\mathbf{\sqrt{x^2+y^2}=\sqrt{\frac{-y}{1+2xy}}}$\\\\\\ So $\mathbf{\mid z \mid =\sqrt{\frac{-y}{1+2xy}}}$\\\\\\ $\mathbf{\mid z \mid=\sqrt{\frac{-\mathbb{I}m(z)}{1+2\mathbb{R}e(z).\mathbb{I}m(z)}}}$
