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Find imaginary part

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Poirot

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Feb 15, 2012
250
Show that,

\[\mbox{Im}(f(z))=\frac{1-|z|^2}{|z-i|^2} \mbox{ where }f(z)=\frac{z+i}{iz+1}\]

\begin{eqnarray}

\mbox{Im}(f(z))&=&\frac{1}{2i}(f(z)-\overline{f(z)})\\

&=&\frac{1}{2i}\left(\frac{z+i}{iz+1}-\frac{\overline{z}-i}{-i\overline{z}+1}\right)\\

&=&\frac{1}{2i}\left(\frac{(z+i)(-i\overline{z}+1)-(iz+1)(\overline{z}-i)}{|iz+1|^2}\right)\\

&=&\frac{1}{2i}\left(\frac{(z+i)(-i\overline{z}+1)-(iz+1)(\overline{z}-i)}{|iz+1|^2}\right)

\end{eqnarray}
 
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ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
\(\displaystyle f(z)=\frac{z+i}{iz+1}\).

We know that \(\displaystyle z\cdot \bar{z} = |z|^2\)

\(\displaystyle f(z)=\frac{z+i}{iz+1}= \frac{1-iz}{z-i}=\frac{(1-iz)(\bar{z}+i)}{|z-i|^2}=\frac{\bar{z}+i+z-i|z|^2}{|z-i|^2}\)

\(\displaystyle \text{Im} \left( \frac{2\text{Re}(z)+i(1-|z|^2)}{|z-i|^2}\right) = \frac{1-|z|^2}{|z-i|^2}\)
 

Prove It

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MHB Math Helper
Jan 26, 2012
1,403
Show that,

\[\mbox{Im}(f(z))=\frac{1-|z|^2}{|z-i|^2} \mbox{ where }f(z)=\frac{z+i}{iz+1}\]

\begin{eqnarray}

\mbox{Im}(f(z))&=&\frac{1}{2i}(f(z)-\overline{f(z)})\\

&=&\frac{1}{2i}\left(\frac{z+i}{iz+1}-\frac{\overline{z}-i}{-i\overline{z}+1}\right)\\

&=&\frac{1}{2i}\left(\frac{(z+i)(-i\overline{z}+1)-(iz+1)(\overline{z}-i)}{|iz+1|^2}\right)\\

&=&\frac{1}{2i}\left(\frac{(z+i)(-i\overline{z}+1)-(iz+1)(\overline{z}-i)}{|iz+1|^2}\right)

\end{eqnarray}
\(\displaystyle \displaystyle \begin{align*} z &= x + i\,y \textrm{ where } x, y \in \mathbf{R} \\ \\ f(z) &= \frac{z + i}{i\,z + 1} \\ &= \frac{x + i\,y + i}{i\left( x + i\,y \right) + 1} \\ &= \frac{x + i \left( 1 + y \right) }{ 1 - y + i\,x } \\ &= \frac{\left[ x + i \left( 1 + y \right) \right] \left( 1 - y - i\,x \right) }{ \left( 1 - y + i\,x \right) \left( 1 - y - i\,x \right) } \\ &= \frac{ x \left( 1 - y \right) - i\, x^2 + i \left( 1 + y \right) \left( 1 - y \right) + x \left( 1 + y \right) }{ \left( 1 - y \right) ^2 + x^2 } \\ &= \frac{2x + i \left( 1 - x^2 - y^2 \right) }{ x^2 + \left( 1 - y \right) ^2 } \\ &= \frac{2x}{ x^2 + \left( 1 - y \right)^2 } + i \left[ \frac{1 - \left( x^2 + y^2 \right) }{ x^2 + \left( 1 - y \right)^2 } \right] \end{align*}\)

So therefore

\(\displaystyle \displaystyle \begin{align*} \mathcal{I} \left[ f(z) \right] &= \frac{1 - \left( x^2 + y^2 \right) }{ x^2 + \left( 1 - y \right) ^2 } \end{align*}\)

And since \(\displaystyle \displaystyle \begin{align*} \left| z \right|^2 = x^2 + y^2 \end{align*}\) and \(\displaystyle \displaystyle \begin{align*} \left| z - i \right| ^2 = x^2 + \left( y - 1 \right) ^2 = x^2 + \left( 1 - y \right) ^2 \end{align*}\) that means

\(\displaystyle \displaystyle \begin{align*} \mathcal{I} \left[ f(z) \right] &= \frac{ 1 - \left| z \right| ^2 }{ \left| z - i \right| ^2 } \end{align*}\)