# Reciprocal of Complex Number?

• Oct 10th 2009, 06:23 AM
Viral
Reciprocal of Complex Number?
$z=\frac{3+4i}{2-3i}$. What is the complex number which satisfies the equation $zw=1$.

My attempt:

$\begin{array}{rcrcrc}
z=\frac{3+4i}{2-3i}\\
\\
zw=1\\
\\
w=\frac{1}{z}\\
\\
\frac{3+4i}{2-3i}*\frac{2+3i}{2+3i}=\frac{-6+17i}{13}
\end{array}$

Is it right up to there? If so, I see I need to get the reciprocal of that (which will be the value of w). How would I do that?
• Oct 10th 2009, 06:25 AM
Matt Westwood
Quote:

Originally Posted by Viral
$z=\frac{3+4i}{2-3i}$. What is the complex number which satisfies the equation $zw=1$.

My attempt:

$\begin{array}{rcrcrc}
z=\frac{3+4i}{2-3i}\\
\\
zw=1\\
\\
w=\frac{1}{z}\\
\\
\frac{3+4i}{2-3i}*\frac{2+3i}{2+3i}=\frac{-6+17i}{13}
\end{array}$

Is it right up to there? If so, I see I need to get the reciprocal of that (which will be the value of w). How would I do that?

My initial utterly naive approach would be to say that $1/(a/b) = b/a$ and so $w = 1/z = \frac{2-3i}{3+4i}$.

Or am I missing something subtle?
• Oct 10th 2009, 06:28 AM
Viral
Ahh, I was trying to do that after working it out >< . I didn't think it would make a difference. Thanks, I'll try that out.