Question

Prove the following inequalities: $\left| \dfrac{z}{\left| z \right|}-1 \right|\le \arg z$

Answer 1

Hint: Put $z=r{{e}^{i\theta }}=r\left( \cos \theta +i\sin \theta \right)$ i.e. the Euler form of complex number in the given inequality, where r is the modulus of the complex number and $\theta $ is the argument of it. Use the following relations to prove the given identity:-
${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ , ${{\cos }^{2}}\theta =1-2{{\sin }^{2}}\theta $
$\sin \theta \le \theta $ , for $\theta \ge 0$ .

Complete step-by-step answer:

So, we need to prove the inequality:-

$\left| \dfrac{z}{\left| z \right|}-1 \right|\le \arg z$ (i)

To prove the above inequality, we need to simplify the L.H.S and R.H.S both of the inequality. So, let us put the Euler form of the complex number which is given as

$z=r{{e}^{i\theta }}=r\left( \cos \theta +i\sin \theta \right)$ (ii)

Where r is the modulas of the complex number ‘z’ and ‘$\theta $’ is the argument of z.

So, we can get L.H.S of the equation (i) as LHS $=\left| \dfrac{z}{\left| z \right|}-1 \right|$

LHS $=\left| \dfrac{r{{e}^{i\theta }}}{r}-1 \right|$

LHS $=\left| {{e}^{i\theta }}-1 \right|=\left| \left( \cos \theta +i\sin \theta \right)-1 \right|$

LHS $\left| \left( \cos \theta -1 \right)+i\sin \theta \right|$ (iii)

As, we know modulas of any complex number $'z'=x+iy$ is given as

$\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$ (iv)

Hence, we can get LHS of equation (iii) from equation (iv) as

$\begin{align}

  & LHS=\sqrt{{{\left( \cos \theta -1 \right)}^{2}}+{{\left( \sin \theta \right)}^{2}}} \\

 & LHS=\sqrt{{{\cos }^{2}}\theta +1-2\cos \theta +{{\sin }^{2}}\theta } \\

\end{align}$

$LHS=\sqrt{{{\cos }^{2}}\theta +{{\sin }^{2}}\theta +1-2\cos \theta }$ (v)

As, we know ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ , So, we can get value of LHS from equation (v) as

$\begin{align}

  & LHS=\sqrt{1+1-2\cos \theta } \\

 & LHS=\sqrt{2-2\cos \theta } \\

\end{align}$

$LHS=\sqrt{2\left( 1-\cos \theta \right)}$ (vi)

We know value of ${{\cos }^{2}}\theta $ can be given as ${{\cos }^{2}}\theta =1-2{{\sin }^{2}}\theta $

Replace $\theta $ by $\dfrac{\theta }{2}$ in the above expression to get value of $\cos \theta $ ; we get

$\cos \theta =1-2{{\sin }^{2}}\dfrac{\theta }{2}$ (vii)

So, put the value of $\cos \theta $ from equation (vii) to equation (vi). Hence, we get

$\begin{align}

  & LHS=\sqrt{2\left( 1-\left( 1-2{{\sin }^{2}}\dfrac{\theta }{2} \right) \right)} \\

 & LHS=\sqrt{2-2+4{{\sin }^{2}}\dfrac{\theta }{2}} \\

 & LHS=\sqrt{4{{\sin }^{2}}\dfrac{\theta }{2}} \\

\end{align}$

$LHS=2\left| \sin \dfrac{\theta }{2} \right|=\pm 2\sin \dfrac{\theta }{2}$ (ix)

Value of RHS of the equation (i) is given as

$\arg z=\arg \left( r{{e}^{i\theta }} \right)=\theta $ (x)

Now, we know the inequality that

$\sin \theta \le \theta $ for all $\theta \ge 0$ (xi)

So, we can replace $\theta $ by $\dfrac{\theta }{2}$ to the above expression, we get

$\begin{align}

  & \sin \dfrac{\theta }{2}\le \dfrac{\theta }{2} \\

 & \Rightarrow 2\sin \dfrac{\theta }{2}\le \dfrac{\theta }{2}\times 2 \\

 & 2\sin \dfrac{\theta }{2}\le \theta \\

\end{align}$

Now we can put modulas sign to the LHS, as the inequality is defined for only positive

$'\theta '$ .So, we get

$2\left| \sin \dfrac{\theta }{2} \right|\le \theta $

Hence, we can replace $2\left| \sin \left( \dfrac{\theta }{2} \right) \right|$ by $\left| \dfrac{z}{\left| z \right|}-1 \right|$ from the equation (i) and equation (ix), and $\theta $ by the $\arg z$from the equation (x). So, we get $\left| \dfrac{z}{\left| z \right|}-1 \right|\le \arg z$.

Hence, the given inequality is proved.


Note: Another approach for the question would be that we can put value of $z=x+iy$ to the given inequality to both sides, where $\arg z={{\tan }^{-1}}\left( \dfrac{y}{x} \right)$ and $\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$ . But it will be a very difficult and complex approach with the involvement of two variables x and y. so, it will give a solution but will take longer time and expression will become very complex.