Reciprocal of Complex Number?

**Viral** · October 10th 2009, 06:23 AM

$z=\frac{3+4i}{2-3i}$ . 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?

**Matt Westwood** · October 10th 2009, 06:25 AM

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?

**Viral** · October 10th 2009, 06:28 AM

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.

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