Prove $\arg(zw) = \arg ((z+w)^2)$ given that $|z| = |w|$

Question

I started with the RHS and only managed to do get the following: $\arg(z^2 + 2zw +w^2) = \arg(z^2) + \arg(w^2) + \arg(2zw)$ $= 2\arg(z) + 2\arg(w) + \arg(zw)$

I'm not sure if what I did in the last line is correct, but I suppose that $\arg(2zw) = \arg(zw)$ as the $2$ only doubles the modulus.

Edit: I see the error in my working out. I'm still not sure how to go about proving it though.

Edit 2: Thanks everyone. I see that a graphical approach is the easiest way to deal with this question.

Answer 1

Well, $$\arg(zw)=\arg((z+w)^2) \iff \frac{\arg(z)+\arg(w)}{2}=\arg(z+w) \tag{1}$$ and from here I would explain this graphically because this just shows how beautiful and useful complex numbers are. (and probably because I can't somehow give a complete proof of it algebraically xD)

If you denote the origin point as $O$, then this means that the point $z+w$ on the argand diagram lies on the angle bisector of $\angle ZOW$, where the point $Z=(a,b)$ when $z=a+bi$ and same with $W$ and $w$.

Now note that when you add two complex numbers you translate one by a vector same as the other. The same thing when you add two vectors. You can look up complex numbers addition graphically, because I can't draw it here.

Now let the point which graphs the complex number $z+w$ on the argand diagram be $P$, we have the quadrilateral $ZOWP$ is a rhombus, since $ZO=OW=WP=PZ$ and the parallelism is easy to be shown when displaying point $P$. Now, since it is a rhombus, $OP$ naturally bisects $\angle ZOW$ which is equivalent to $(1)$.

Note: $(1)$ is concluded from the fact that for any two complex numbers $a$ and $b$, $\arg(ab)=\arg(a)+\arg(b)$

Answer 2

• If $z=\pm w,$ the proof is straightforward. Use the property mentionned in comments: $$\arg(z_1z_2)=\arg z_1 + \arg z_2.$$

• Assume that $z\neq \pm w.$

The assumption $|z|=|w|$ is fundamental: the points with affices $\;0,z,z+w, w\;$ are vertices of a rhombus. Diagonals of a rhombus are also angle bissectors. Therefore, $\arg (z+w)={1\over2}(\arg z + \arg w).$

With the use of the above mentionned property, we obtain $$LHS=\arg (zw) = \arg z + \arg w$$ $$RHS=\arg((z+w)^2)=\arg(z+w)+\arg(z+w)=2\cdot {1\over2}(\arg z + \arg w),$$ which finishes the proof.

Answer 3

You have to use $$ \arg(z_1z_2)=\arg(z_1)+\arg(z_2)\mod(-\pi,\pi] $$ and $$ \arg(z)=\arctan(\frac{b}{a}) \mod(-\pi,\pi]$$ for $z=a+bi$. Let $z=re^{\theta i},w=re^{i\omega}$. Then \begin{eqnarray} (z+w)^2&=&r^2(e^{2\theta i}+e^{2\omega i}+2e^{(\theta+\omega)i}\\ &=&r^2\bigg[(\cos(2\theta)+\cos(2\omega)+2\cos(\theta+\omega))+i(\sin(2\theta)+\sin(2\omega)+2\sin(\theta+\omega))\bigg]\\ &=&r^2\bigg[(2\cos(\theta+\omega)\cos(\theta-\omega)+2\cos(\theta+\omega))+i(2\sin(\theta+\omega)\cos(\theta-\omega)+2\sin(\theta+\omega))\bigg]\\ &=&r^2\bigg[2\cos(\theta+\omega)(\cos(\theta-\omega)+1)+2i\sin(\theta+\omega)(\cos(\theta-\omega)+1)\bigg] \end{eqnarray} and hence $$ \arg((z+w)^2)=\arctan\bigg(\frac{\sin(\theta+\omega)(\cos(\theta-\omega)+1)}{2\cos(\theta+\omega)(\cos(\theta-\omega)+1)}\bigg)=\arctan(\tan(\theta+\omega))=\theta+\omega=\arg(zw)\mod(-\pi,\pi]. $$

Answer 4

We know $\arg z = \arg w$ if and only if $z/w$ is real, i.e. $\Im(z/w) = 0$. So we would like to show that $$\Im \left( \frac{(z+w)^2}{zw} \right) = 0$$ which is equivalent to $$\frac{(z+w)^2}{zw} - \frac{(\bar z+ \bar w)^2}{\bar z \bar w} = 0$$ which is equivalent to $$\bar z \bar w (z+w)^2 - zw (\bar z + \bar w)^2 = 0$$ which is an easy identity, using the information that $z \bar z = w \bar w$.