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I want to calculate the absolute value and argument of the complex number $a = \left(\sqrt{3} - i\right)^{-2}$.

In order to calculate these two values I tried to reform the number into the form $z = x + y \cdot i$:

$$a =\left(\sqrt{3} - i\right)^{-2} = \frac{1}{\left(\sqrt{3}-i\right)^2} = \frac{1}{2 +\left(-2\sqrt{3}\right)}$$

$$ = -\frac{2}{8} - i \frac{-2\sqrt{3}i}{4-12} = -\frac{1}{4} - i \frac{\sqrt{3}i}{4}$$

I now want to calculate the polar form of a:

$$r = |z| = \sqrt{x^2 + y^2} = \sqrt{\frac{1}{8} + \left( -1 -2i \left( \frac{\sqrt{3}i}{4}\right) + \left(\frac{\sqrt{3}i}{4}\right)^2 \right)}$$ $$ = \sqrt{\frac{1}{8} + \left( -1 -2i \left( \frac{\sqrt{3}i}{4}\right) + \frac{3 + 2 \sqrt{3}i - 1}{16} \right)} = \sqrt{\frac{1}{8} + \left( -1 -i \left( \frac{\sqrt{3}i}{2}\right) + \frac{3 + 2 \sqrt{3}i - 1}{16} \right)} $$ $$=\sqrt{\frac{1}{8} + -1 \frac{\sqrt{3}-1}{2} + \frac{2 + 2 \sqrt{3}i}{16}}$$

but finally am stuck here. Can you please help me to go on? Is there a better way to get to the absolute value and argument?

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    $\begingroup$ There's a much easier way to get the absolute value of this complex number: $|z|^2=zz^*=\frac{1}{(\sqrt3-i)^2}\frac{1}{(\sqrt3+i)^2}=\frac{1}{(3+1)^2}$. $\endgroup$ – David H Apr 21 '14 at 18:09
  • $\begingroup$ @muffel Or how about DeMoivre ? $\endgroup$ – imranfat Apr 21 '14 at 18:20
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I would start by finding the polar form of $z = \sqrt{3} - i$. This corresponds to a nice triangle, the so called 30-60-90 triangle. Here one leg is of length $\sqrt{3}$ and the other is of length 1 (in the negative imaginary direction). The hypotenuse of this triangle is the magnitude of $z$, so $|z|=2$.

The argument of this $z$ is $\theta = -\pi/6$.

Now the number you want is $$a = 1/z^2 = \frac{1}{|z|^2e^{i2\theta}} = \frac{1}{|z|^2} e^{-2i\theta}.$$

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  • $\begingroup$ For questions like this, the choice of components corresponding to a standard triangle is not a coincidence. These are some of the few choices where we can write the argument explicitly using our knowledge of trigonometry. $\endgroup$ – Joel Apr 21 '14 at 18:33
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Rewrite: \begin{align} (\sqrt{3}-i)^{-2}&=\frac{1}{(\sqrt{3}-i)^{2}}\\ &=\frac{1}{2-2\sqrt{3}\ i}\\ &=\frac{1}{2-2\sqrt{3}\ i}\cdot\frac{2+2\sqrt{3}\ i}{2+2\sqrt{3}\ i}\\ &=\frac{2+2\sqrt{3}\ i}{2^2-(2\sqrt{3}\ i)^2}\\ &=\frac{2+2\sqrt{3}\ i}{16}\\ &=\frac18+\frac18\sqrt{3}\ i \end{align} It should be easy from this to obtain its absolute value and argument. The rest, I leave it to you. Good luck! :)


It looks like you didn't know to evaluate the absolute value and argument of $|z|$. The absolute value can be obtained by using this formula $$ |z|=\sqrt{x^2+y^2} $$ and the argument of $z$ is $$ \arg(z)=\arctan\left(\frac yx\right). $$ In this case, we have $x=\Re(z)=\frac18$ and $y=\Im(z)=\frac18\sqrt{3}$. You may refer to this link to learn more about complex number. I hope this helps.

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