# Equality of complex numbers in general

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.

• Note that the statement is false when $z_1=z_2=0$. – Greg Martin Jul 20 at 16:53

\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}

• Brilliant! thanks a lot. – S.H.W Jul 20 at 7:54
• 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. – TheSimpliFire Jul 20 at 7:59
• Yes, that is right. – S.H.W Jul 20 at 8:02
• 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.) – Greg Martin Jul 20 at 16:55
• @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. – TheSimpliFire Jul 20 at 18:08

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 $$\theta_1=\theta_2+2k\pi$$.

• Why $z_1 = z_2$ implies $r_1=|z_1|=|z_2|=r_2$? – S.H.W Jul 20 at 7:37
• @S.H.W I've edit my answer to make it more clear. Can you follow now? – Feng Shao Jul 20 at 7:41
• Yes, thanks for the answer. – S.H.W Jul 20 at 7:57

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?

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

If $$\cos\theta_1=\cos\theta_2$$ and $$\sin\theta_1=\sin\theta_2$$

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

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

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 $$\sin^2\dfrac{\theta_1+\theta_2}2+\cos^2\dfrac{\theta_1+\theta_2}2=1$$

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 $$2\pi|\theta_1-\theta_2$$.