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I have trouble algebraically show proof for this well known statement:

$$-\operatorname{arg}(z)=\operatorname{arg}(z^{-1})=\operatorname{arg}(\bar{z})$$

if $z=x+yi$ and $-\operatorname{arg}(z)=-\arctan(y/x)$ right? This one I understood if $\arctan(-\phi)=-\arctan(\phi)$

$$\begin{align} -\arg (z)&=\arg(\bar{z}) \\ -\arctan(y/x)&=\arctan(-y/x) \\ -\arctan(y/x)&=-\arctan(y/x)\end{align}$$

But I have trouble with showing that $-\operatorname{arg}(z)=\operatorname{arg}(z^{-1})$ or is it something like this. $$1/z=\frac {1}{x+iy}=\frac{x-iy}{x^2+y^2}$$ EDIT: I think I understood.. $$-\arg(z)=\arg(1/z)=\arctan \Big( \frac{\dfrac{-y}{x^2+y^2}}{\dfrac{x}{x^2 +y^2}} \Big)=\arctan(-y/x)=-\arg(z) $$

Did I get it right?

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Using the inverse tangent function $\arctan$ to find the argument only works if you restrict yourself to two adjacent quadrants. For example, $1+\mathrm{i}$ and $\mathrm{-1}-\mathrm{i}$ both have $$\frac{y}{x}=1$$ and so would have the same argument if $\arg(z) = \arctan\!\left(\frac{y}{x}\right)$.

Similarily, $-1+\mathrm{i}$ and $1-\mathrm{i}$ both have $\frac{y}{x}=-1$ and would have the same argument.

I would recommend that you use the polar form $z=r(\cos \theta + \mathrm{i}\sin\theta)$, where $r$ is the modulus of $z$ and $\theta$ is the argument. The sine and cosine functions repeat every $2\pi$. We have:

$$\frac{1}{z} = \frac{1}{r(\cos\theta+\mathrm{i}\sin\theta)} = \frac{1}{r}\cdot\frac{1}{\cos\theta+\mathrm{i}\sin\theta}$$ We multiply numerator and denominator by the conjugate: $$\begin{eqnarray*} \frac{1}{z} &=& \frac{1}{r}\cdot\frac{1}{\cos\theta+\mathrm{i}\sin\theta}\cdot\frac{\cos\theta-\mathrm{i}\sin\theta}{\cos\theta-\mathrm{i}\sin\theta} \\ \\ &=&\frac{1}{r}\cdot\frac{\cos\theta+\mathrm{i}\sin\theta}{\cos^2\theta+\sin^2\theta} \\ \\ &=& \frac{1}{r}\cdot (\cos\theta-\mathrm{i}\sin\theta) \end{eqnarray*}$$ Two well-known identities are: $\cos(-\theta) \equiv \cos \theta$ and $\sin(-\theta) \equiv - \sin\theta$. It follows that $$\frac{1}{z} = \frac{1}{r}\cdot (\cos\theta-\mathrm{i}\sin\theta) = \frac{1}{r}\cdot(\cos(-\theta)+\mathrm{i}\sin(-\theta))$$ It follows that $$\begin{eqnarray*} \left| \frac{1}{z} \right| &=& \frac{1}{|z|} \\ \\ \arg\left(\frac{1}{z}\right) &=& -\arg z \end{eqnarray*}$$

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  • $\begingroup$ Ahh yeah THANKS! The statement $\bar{z}$ reduces to $r(\cos \theta -i \sin \theta )$ similarly and of with those sine and cosine identities we get the same result! $\endgroup$ – ELEC Sep 5 '14 at 16:16
  • $\begingroup$ Does the identity $\arg \bar{z} = -\arg z$ hold for all complex numbers by the way? $\endgroup$ – ELEC Sep 5 '14 at 16:29
  • $\begingroup$ @ELEC You're very welcome. The short answer is yes: $\arg\overline{z} = -\arg z$ for all $z \neq 0$. The argument of $0$ is not defined. Long answer: We need to be careful though because the argument is not unique. For example: $\arg(1) = 0, \pm 2\pi, \pm 4 \pi, \ldots$. We need to use the so-called principal argument, i.e. $-\pi < \arg z \le \pi$. That has $\arg\overline{z} = -\arg z$ for all non-positive real numbers. The argument of $z=0$ is not defined and the arguments of the negative reals are all $\pi$. $\endgroup$ – Fly by Night Sep 5 '14 at 17:57
  • $\begingroup$ @ELEC See here for more details: en.wikipedia.org/wiki/… $\endgroup$ – Fly by Night Sep 5 '14 at 18:01
  • $\begingroup$ Thanks a lot for a thorough answer @FlyByNight! I learned so much new stuff. $\endgroup$ – ELEC Sep 6 '14 at 7:40
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So, because we're only considering the argument we can stay on the unit circle. Then, if $z = e^{i \theta}$ then $z^{-1} = e^{-i\theta} = \overline{z}$. Moreover \begin{equation} -\operatorname{arg}(z) = -\theta = \arg(z^{-1}) = \arg(\overline{z}) \end{equation}

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  • $\begingroup$ I'm not sure that your assumption is automatic. We cannot assume that the argument of $\frac{1}{z}$ is independent of the modulus of $z$. A consequence of the proof that $\arg(1/z)=-\arg(z)$ is that the argument of $1/z$ is independent of the modulus of $z$. $\endgroup$ – Fly by Night Sep 5 '14 at 15:49
  • $\begingroup$ I'm fairly sure it is automatic as we define the modulus and argument to be the distance from zero and angle of elevation respectively. If nothing else, the arctan form of the argument quite easily shows that the only thing that affects the argument is the ratio of the real and imaginary parts, not the absolute value. $\endgroup$ – Stijn Hanson Sep 5 '14 at 16:03
  • $\begingroup$ You need to add these explanations to your answer. If the OP knew what you know then s/he wouldn't be asking the question. A good answer justifies any assumptions. $\endgroup$ – Fly by Night Sep 5 '14 at 16:09
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Since you are talking about a non-zero complex number $z$, you should know that $z$ can always be written as $z=re^{i\theta}$ with $r>0$ and $\theta\in \mathbb R$.

This is a polar expression of $z$ and any other polar expression of $z$ is of the form $\rho e^{i\psi}$ where $\rho=r$ and $\psi=\theta+2k\pi$, $k\in \mathbb Z$.

Now, by definition, $arg(z)$ is given by $\theta$ up to $2\pi$ that is : $$arg(z)=\theta \pmod {2\pi}.$$

If the above explanation (or definition) is clear to you, you can now compute anything you want.

Observe that $$\frac{1}{z}=\frac{1}{r}e^{-i\theta}$$ and this is a polar expression so $$\begin{array}{rl} arg(z^{-1}) & =-\theta \pmod {2\pi} \\ & = -arg(z)\end{array}$$ Similarly, $$\bar{z}=\overline{re^{i\theta}} =\bar{r}\cdot \overline{e^{i\theta}}=re^{-i\theta}$$ which is a polar form, so $$\begin{array}{rl} arg(\bar z) & =-\theta \pmod {2\pi} \\ & = -arg(z)\end{array}$$

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