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$$ \frac{1}{(x+iy)^2}=\frac{1}{x^2+i2xy-y^2}=\frac{x^2}{(x^2+y^2)}-\frac{2ixy}{(x^2+y^2)}-\frac{y^2}{(x^2+y^2)}$$ So I thought I could just say: $$ Re(\frac{1}{z^2})=\frac{x^2}{(x^2+y^2)}-\frac{y^2}{(x^2+y^2)}$$ and $$ Im(\frac{1}{z^2})=-\frac{2ixy}{(x^2+y^2)}$$ But I know that is wrong because it looks nothing like the graph of the real part of 1/z^2 on wolfram alpha found here: http://www.wolframalpha.com/input/?i=1%2F(x%2Bi*y)%5E2

Then I thought I could must multiply $1/z^2$ by $z/z$ to get $\frac{x}{z^3}$ and $\frac{iy}{z^3}$ however graphing these again shows that they are not the real and complex parts of $\frac{1}{z^2}$.

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    $\begingroup$ Look carefully at the denominator. $(x^2+y^2)$ should be squared. $\endgroup$ – Oscar Lanzi Aug 12 '16 at 1:33
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    $\begingroup$ Small but important point - the Imaginary part $Im$ function returns a real value. You shouldn't have the $i$ in your expression. (In addition to the error you are asking about.) $\endgroup$ – Ian Miller Aug 12 '16 at 1:33
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Notice, when $z\in\mathbb{C}$:

$$z=\Re[z]+\Im[z]i$$


So, we get (in steps):

  • $$z^2=\left(\Re[z]+\Im[z]i\right)^2=\Re^2[z]-\Im^2[z]+2\Re[z]\Im[z]i$$
  • $$\overline{z^2}=\overline{\Re^2[z]-\Im^2[z]+2\Re[z]\Im[z]i}=\Re^2[z]-\Im^2[z]-2\Re[z]\Im[z]i$$
  • $$z^2\cdot\overline{z^2}=|z|^4=\left(\sqrt{\Re^2[z]+\Im^2[z]}\right)^4=\left(\Re^2[z]+\Im^2[z]\right)^2$$

Now, we get:

$$\frac{1}{z^2}=\frac{\overline{z^2}}{z^2\cdot\overline{z^2}}=\frac{\Re^2[z]-\Im^2[z]-2\Re[z]\Im[z]i}{\left(\Re^2[z]+\Im^2[z]\right)^2}$$

So:

  • $$\color{red}{\Re\left[\frac{1}{z^2}\right]=\frac{\Re^2[z]-\Im^2[z]}{\left(\Re^2[z]+\Im^2[z]\right)^2}}$$
  • $$\color{red}{\Im\left[\frac{1}{z^2}\right]=-\frac{2\Re[z]\Im[z]}{\left(\Re^2[z]+\Im^2[z]\right)^2}}$$
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Your first proposition should be corrected by noting that when you multiply numerator and denominator by the conjugate of $x^2+2ixy-y^2$, the denominator becomes $(x+iy)^2(x-iy)^2=(x^2+y^2)^2$ and not $x^2+y^2$. Moreover, there's no $i$ in the imaginary part.

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  • $\begingroup$ $ Re(\frac{1}{z^2})=\frac{x^2}{(x^2+y^2)^2}-\frac{y^2}{(x^2+y^2)^2}$ and $ Im(\frac{1}{z^2})=-\frac{2ixy}{(x^2+y^2)^2}$ $\endgroup$ – User3910 Aug 19 '16 at 23:15
  • $\begingroup$ That's right... $\endgroup$ – paf Aug 19 '16 at 23:20
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\begin{eqnarray} f(z)=1/(z^2)=1/((x+iy)^2)&=&1/(x^2-y^2+2ixy) \newline &=&(1/(x^2-y^2+2ixy))\cdot((x^2-y^2-2ixy)/(x^2- y^2-2ixy)) \newline &=&((x^2-y^2-2ixy)/((x^2-y^2)^2+4x^2y^2)) \newline &=&(x^2-y^2-2ixy)/(x^2+y^2)^2 \end{eqnarray} therefore $\Re(1/z^2)=(x^2-y^2)/(x^2+y^2)^2$ and $\Im(1/z^2)=-2xy/(x^2+y^2)^2.$

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  • $\begingroup$ Please use MathJax to format expressions. $\endgroup$ – Saad Jan 11 '18 at 5:08

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