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How does $(iwl + \frac{1}{iwc})^2$ equal to $(wl - \frac{1}{wc})^2$?

Let me clarify. In physics there is the impedance which is a complex number
Z = R + iwl + 1/iwc

R, w, l, and c, are constants; i = $\sqrt-1$
According to my book: |z| = $\sqrt{R^2 + (wl - \frac{1}{wc})^2}$

The real part of Z (i.e. R) squared gives R^2; the imaginary part squared gives the second term under the root, but I don't get how they're equivalent.

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They're not:

$$\left( iwl + {1 \over iwc} \right)^2 = \left( iwl - {i \over wc} \right)^2 = i^2 \left( wl - {1 \over wc} \right)^2 = - \left( wl - {1 \over wc} \right)^2$$

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  • $\begingroup$ I clarified the question $\endgroup$ – user35687 Jan 20 '15 at 21:05
  • $\begingroup$ In which case yes, because for any complex number $z$ can be written as $z = a + bi$ for $a, b$ real, and has modulus $z = \sqrt{a^2 + b^2}$. Now apply that formula to your $Z$. $\endgroup$ – Simon S Jan 20 '15 at 21:40

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