- #1
songoku
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- Homework Statement
- For ##z+\frac 1 z = 1+ 2i##, find ##|z^2 - \frac{1}{z^2}|##
- Relevant Equations
- conjugate
modulus of complex number: ##|z|=\sqrt{x^2+y^2}##
Let: ##z=x+iy##
$$z+\frac 1 z =1+2i$$
$$x+iy +\frac{1}{x+iy}=1+2i$$
$$x+iy+\frac{1}{x+iy} . \frac{x-iy}{x-iy}=1+2i$$
$$x+iy+\frac{x-iy}{x^2+y^2}=1+2i$$
$$\frac{x^3+xy^2+x+i(x^2y+y^3-y)}{x^2+y^2}=1+2i$$
So:
$$\frac{x^3+xy^2+x}{x^2+y^2}=1$$
$$x^3+xy^2+x=x^2+y^2$$
and
$$\frac{x^2y+y^3-y}{x^2+y^2}=2$$
$$x^2y+y^3-y=2(x^2+y^2)$$
Combining the two equations: ##x^2y+y^3-y=2(x^3+xy^2+x)##
How to solve that equation? Or maybe there is another approach to this question?
Thanks
$$z+\frac 1 z =1+2i$$
$$x+iy +\frac{1}{x+iy}=1+2i$$
$$x+iy+\frac{1}{x+iy} . \frac{x-iy}{x-iy}=1+2i$$
$$x+iy+\frac{x-iy}{x^2+y^2}=1+2i$$
$$\frac{x^3+xy^2+x+i(x^2y+y^3-y)}{x^2+y^2}=1+2i$$
So:
$$\frac{x^3+xy^2+x}{x^2+y^2}=1$$
$$x^3+xy^2+x=x^2+y^2$$
and
$$\frac{x^2y+y^3-y}{x^2+y^2}=2$$
$$x^2y+y^3-y=2(x^2+y^2)$$
Combining the two equations: ##x^2y+y^3-y=2(x^3+xy^2+x)##
How to solve that equation? Or maybe there is another approach to this question?
Thanks