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Suppose $z_1 = r_1(\cos \theta_1 + i\sin \theta_1)$ and $z_2 = r_2(\cos \theta_2 + i\sin \theta_2)$. Prove that $z_1 = z_2$ $\iff$ $r_1 = r_2 , \theta_1 = \theta_2 + 2k\pi$ .

My try: The proof for the converse statement is obvious but I don't know how to prove the forward statement. I tried to use "Two complex numbers $a+bi$ and $c+di$ are equal iff $a=b$ and $c=d$, where $a,b,c,d$ are real numbers." but it didn't work.

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    $\begingroup$ Note that the statement is false when $z_1=z_2=0$. $\endgroup$ – Greg Martin Jul 20 at 16:53
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\begin{align}z_1=z_2&\implies r_1\cos\theta_1+ir_1\sin\theta_1=r_2\cos\theta_2+ir_2\sin\theta_2\\&\implies\begin{cases}r_1\cos\theta_1=r_2\cos\theta_2\\r_1\sin\theta_1=r_2\sin\theta_2\end{cases}\\&\implies \tan\theta_1=\tan\theta_2\quad (r_1,r_2>0\,\land\,\cos\theta_1,\cos\theta_2\ne0 \,\,{}^{**})\\&\implies\theta_1=\theta_2+K\pi\quad(K\in\Bbb Z)\\&\implies r_2\cos\theta_2=r_1\cos(\theta_2+K\pi)=r_1\cos\theta_2\cos K\pi-r_1\sin\theta_2\sin K\pi\\&\implies r_2\cos\theta_2=r_1\cos\theta_2\cos K\pi\\&\implies r_2=r_1\cos K\pi\quad(\cos\theta_2\ne0)\\&\implies r_2=r_1\quad(\cos K\pi=\pm1;\,r_1,r_2>0\implies\cos K\pi=1\implies K\equiv0\pmod2)\\&\implies r_1=r_2\,\land\, \theta_1=\theta_2+2k\pi\quad(k\in\Bbb Z)\end{align} ${}^{**}$

\begin{align}\cos\theta_1,\cos\theta_2=0&\implies\theta_1,\theta_2=\frac\pi2+K\pi\quad(K\in\Bbb Z)\\&\implies\sin\theta_1,\sin\theta_2=\pm1\\\cos\theta_1,\cos\theta_2=0&\implies r_1(0+i\sin\theta_1)=r_2(0+i\sin\theta_2)\\&\implies r_1\sin\theta_1=r_2\sin\theta_2\\r_1,r_2>0&\implies\sin\theta_1=\sin\theta_2=1\\&\implies\theta_1=\theta_2+2k\pi\quad(k\in\Bbb Z)\end{align}

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  • $\begingroup$ Brilliant! thanks a lot. $\endgroup$ – S.H.W Jul 20 at 7:54
  • $\begingroup$ Glad it helped, you can try doing a similar thing for $\cos\theta_1=\cos\theta_2=0$ and the same results should be reached. $\endgroup$ – TheSimpliFire Jul 20 at 7:59
  • $\begingroup$ Yes, that is right. $\endgroup$ – S.H.W Jul 20 at 8:02
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    $\begingroup$ Note that the statement is false when $z_1=z_2=0$; and indeed, your proof never returns to the case that you excluded in the third line. (Also, I think that this is a poor writing style to demonstrate for a newer mathematician: someone who didn't already solve the original question will have a hard time understanding the logical structure of this string of math symbols.) $\endgroup$ – Greg Martin Jul 20 at 16:55
  • $\begingroup$ @GregMartin 1) I stated that $r_1,r_2>0$ so the case wouldn't occur. 2) In my comment above I suggested precisely that to the OP, that they can try doing $\cos\theta_i=0$ using a similar method. $\endgroup$ – TheSimpliFire Jul 20 at 18:08
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Note that if $z=r(\cos\theta+i\sin\theta)(r\ge0)$ then $|z|=r$. In fact, $|z|^2=(r\cos\theta)^2+(r\sin\theta)^2=r^2$, then $|z|=r$.

Here I assume $r_1,r_2>0$. Otherwise $z=0=0\cdot(\cos\theta+i\sin\theta)$ holds for every $\theta\in\mathbb R$, in which case the claim in OP's problem is wrong.

$z_1=z_2$ implies that $r_1=|z_1|=|z_2|=r_2$, so $\cos\theta_1=\cos\theta_2$ and $\sin\theta_1=\sin\theta_2$, which means $\theta_1=\theta_2+2k\pi$.

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  • $\begingroup$ Why $z_1 = z_2$ implies $r_1=|z_1|=|z_2|=r_2$? $\endgroup$ – S.H.W Jul 20 at 7:37
  • $\begingroup$ @S.H.W I've edit my answer to make it more clear. Can you follow now? $\endgroup$ – Feng Shao Jul 20 at 7:41
  • $\begingroup$ Yes, thanks for the answer. $\endgroup$ – S.H.W Jul 20 at 7:57
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I think you meant that $r_i\geq0$ and $\theta_i\in\mathbb R,$ otherwise your statement is wrong.

Now, by your work we obtain: $$(r_1\cos\theta_1)^2+(r_1\sin\theta_1)^2=(r_2\cos\theta_2)^2+(r_2\sin\theta_1)^2$$ or $$r_1^2=r_2^2,$$ which gives $$r_1=r_2.$$ Can you end it now?

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  • $\begingroup$ Okay and how $\cos\theta_1=\cos\theta_2 , \sin\theta_1=\sin\theta_2$ implies $\theta_1=\theta_2+2k\pi$? $\endgroup$ – S.H.W Jul 20 at 7:46
  • $\begingroup$ @S.H.W Yes, of course! $\endgroup$ – Michael Rozenberg Jul 20 at 7:48
  • $\begingroup$ I don't know how to prove that. $\endgroup$ – S.H.W Jul 20 at 7:51
  • $\begingroup$ @S.H.W D Draw the trigonometric circle and use the definition of $\sin$ and $\cos$. See here: en.wikipedia.org/wiki/Trigonometric_functions $\endgroup$ – Michael Rozenberg Jul 20 at 8:11
  • $\begingroup$ If $\cos\theta_1=\cos\theta_2$ then $\theta_1 = 2k\pi \pm \theta_2$. If we put this in $\sin\theta_1=\sin\theta_2$ then $\sin\theta_2 = 0$ and $\theta_2 = k\pi$. What is the mistake in this calculation? $\endgroup$ – S.H.W Jul 20 at 8:39
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If $\cos\theta_1=\cos\theta_2$ and $\sin\theta_1=\sin\theta_2$

Using Prosthaphaeresis Formulas,

$$0=\cos\theta_1-\cos\theta_2=-2\sin\dfrac{\theta_1-\theta_2}2\sin\dfrac{\theta_1+\theta_2}2$$

$$0=\sin\theta_1-\sin\theta_2=2\sin\dfrac{\theta_1-\theta_2}2\cos\dfrac{\theta_1+\theta_2}2$$

If $\sin\dfrac{\theta_1-\theta_2}2=0, \dfrac{\theta_1-\theta_2}2$ must be multiple of $\pi$ and we are done.

Else $\sin\dfrac{\theta_1+\theta_2}2=\cos\dfrac{\theta_1+\theta_2}2=0$ which is untenable as $$\sin^2\dfrac{\theta_1+\theta_2}2+\cos^2\dfrac{\theta_1+\theta_2}2=1$$

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Since $z_1^\ast z_1=z_2^\ast z_2$, $r_1^2=r_2^2$ so $r_1=r_2$. Thus $\cos\theta+i\sin\theta_1=\cos\theta_2+i\sin\theta_2$. Dividing by the unit complex number on the right-hand side, $\cos(\theta_1-\theta_2)+i\sin(\theta_1-\theta_2)=1$, i.e. $\cos(\theta_1-\theta_2)=1,\,\sin(\theta_1-\theta_2)=0$. Hence $2\pi|\theta_1-\theta_2$.

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