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If two complex numbers are equal , is it necessary that their arguments are also equal ? Is the vice versa also true ? means that if the arguments of two complex numbers are equal , does it necessarily imply that they’re equal? According to me , the first supposition would be right , but I’m not quite sure about the second one .

Aditi
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  • Hint: what is the argument of any positive real number? – lulu Jan 14 '18 at 18:55
  • @lulu It’s $tan^{-1}|\frac{b}{a}|$ – Aditi Jan 14 '18 at 18:57
  • Neither $a$ nor $b$ appeared in my comment. What is the argument of $3$? What about $39$? – lulu Jan 14 '18 at 18:59
  • Oh for real numbers the argument is $0$ because they lie on the real axis . – Aditi Jan 14 '18 at 19:00
  • Well, for positive real numbers the argument is $0$. Does that tell you what you need to know? – lulu Jan 14 '18 at 19:00
  • @lulu so you mean that it’s not necessary that if two complex numbers are equal they have the same argument ? – Aditi Jan 14 '18 at 19:01
  • Not sure what you are saying, Obviously equal numbers have equal arguments. However my example gives two different complex numbers, $3,39$ as it happens, that have the same argument. Nothing special about those numbers...any two positive reals would have worked. – lulu Jan 14 '18 at 19:04
  • @lulu sorry , I get your point now , that it’s not nexessary that if two complex numbers have the same argument then they’re equal. :) – Aditi Jan 14 '18 at 19:08
  • Yes, that is correct. – lulu Jan 14 '18 at 19:10
  • @lulu thanks for clarifying ! – Aditi Jan 14 '18 at 19:18

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If we take complex numbers as $z_1 = a+bi$ and $z_2 = c+di$, $z_1 = z_2$ if and only if $a = c$ and $b = d$. So $\arg(z_1) = \arg(z_2)$, obviously.

However, if $\arg(z_1) = \arg(z_2)$, it doesn't have to imply that $z_1 = z_2$. A simple counter-example is $z_1 = 2z_2$ with $z_1 \ne 0$. Their arguments are equal however $z_1 \ne z_2$ (if you know some about physics, you can think $z_1$ and $z_2$ as two vectors with same direction but different magnitudes. Then of course we can't say that they are equal).

ArsenBerk
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