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Let's suppose that we have a complex number with

$$r=1 \implies z=\cos\phi+i\sin\phi$$

Then why is $$z^{-1} =\cos (-\phi)+i\sin (-\phi)=\cos\phi-i\sin\phi$$

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    $\begingroup$ Hint: multiply $\cos\varphi+i\sin\varphi$ with $\cos\varphi-i\sin\varphi$ $\endgroup$ – Wojowu Mar 11 '18 at 14:29
  • $\begingroup$ Because the general formula for $z^{-1}$ applied to this case yields this number, because you multiply and see that it makes $1$... Choose your favourite. $\endgroup$ – user228113 Mar 11 '18 at 14:33
  • $\begingroup$ Are you familiar with De Moivre’s theorem? $\endgroup$ – gen-z ready to perish Mar 11 '18 at 17:03
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This is De Moivre's formula in action: For complex $z$ and any integer $n$ we have \begin{align*} z^n=(\cos (\varphi)+i\sin(\varphi))^n=\cos(n\varphi)+i\sin(n\varphi) \end{align*}

The formula \begin{align*} z^{-1}=\cos(-\varphi)+i\sin(-\varphi) \end{align*} is De Moivre's formula evaluated at $n=-1$.

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One way to see is by exponential form

$$z=re^{i\theta}\implies re^{i\theta}\cdot \frac1r e^{-i\theta}=1$$

then

$$z^{-1}=\frac1r e^{-i\theta}=\frac1r(\cos \theta -i\sin \theta) $$

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write $$\frac{1}{\sin(\phi)+i\cos(\phi)}=\frac{\cos(\phi)-i\sin(\phi)}{(\cos(\phi)+i\sin(\phi))(\cos(\phi)-i\sin(\phi))}$$

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Remembering that $z\bar z= |z|^{2}$ imply in this case that $z^{-1}=\bar z$, since $r=\sqrt{|z|^2}$, thus showing that $z^{-1}= \overline {cos(\varphi) +i sin(\varphi)} = cos(-\varphi)+isin(-\varphi)$.

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Note that sin (-φ) = -sin (φ) and cos (-φ) = cos (φ)

Thus $$(\cos (-φ)+i\sin (-φ))(\cosφ-i\sinφ)=$$

$$(cos (-φ)\cosφ -\sin (-φ)\sinφ )+ i(sin (-φ)cosφ-cos (-φ)sinφ)=$$ $$(cos^2φ + sin^2φ) + 0i =1$$

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As your complex number as $r=1$, you can express it like $z=e^{i\theta}$, where $\theta$ is the argument. Then, $$z^{-1}=\frac{1}{z}=\frac{1}{e^{i\theta}}=e^{-i\theta}$$ Now, using the trigonometric form of complex numbers, $$e^{-i\theta}=\cos(-\theta)+i\sin(-\theta)=\cos(\theta)-i\sin(\theta),$$ where we used that $\cos(\theta)=\cos(-\theta)$ and $\sin(\theta)=-\sin(-\theta)$ for all $\theta$.

Edit:

The power series of the exponential, the sine and the cosine are $$e^x=\sum_{n=0}^\infty \dfrac{x^n}{n!}=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}+...$$

$$\sin x=\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n+1)!}x^{2n+1}=x-\dfrac{x^3}{3!}+\dfrac{x^5}{5!}-\dfrac{x^7}{7!}+\dfrac{x^9}{9!}+...$$

$$\cos x=\sum_{n=0}^\infty \dfrac{(-1)^n}{(2n)!}x^{2n}=1-\dfrac{x^2}{2!}+\dfrac{x^4}{4!}-\dfrac{x^6}{6!}+\dfrac{x^8}{8!}+...$$

Now, substituting $i\theta$ instead of $x$ in the power series of $e^x$, we have \begin{align*} e^{i\theta} & =1+i\theta+\dfrac{(i\theta)^2}{2!}+\dfrac{(i\theta)^3}{3!}+\dfrac{(i\theta)^4}{4!}+...=1+i\theta+\dfrac{i^2\theta^2}{2!}+\dfrac{i^3\theta^3}{3!}+\dfrac{i^4\theta^4}{4!}+... \\ \\ & =1+i\theta-\dfrac{\theta^2}{2!}-\dfrac{i\theta^3}{3!}+\dfrac{\theta^4}{4!}+...=\left(1-\dfrac{\theta^2}{2!}+\dfrac{\theta^4}{4!}-...\right)+i\left(\theta-\dfrac{i\theta^3}{3!}+\dfrac{\theta^5}{5!}-...\right) \\ \\ & =\cos\theta +i\sin\theta \end{align*}

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  • $\begingroup$ Why can you express it like $z=e^{iφ)$ $\endgroup$ – Stallmp Mar 11 '18 at 14:50
  • $\begingroup$ All the complex numbers $z=x+iy$ can be expressed as $z=re^{i\theta}$, where $r=\sqrt{x^2+y^2}$ and $\theta=\arctan(\tfrac{y}{x})$. I'll edit my answer proving that. $\endgroup$ – user326159 Mar 11 '18 at 15:03
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Remember: You multiply two complex numbers by adding their angles and multiply their lengths. Now if some two numbers are both $1$ long, just add their angles.

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