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I can't figure this out, it's probably very simple but I just can't seem to solve it. All help is appreciated. $$|z + 1| = |\bar{z} − 1|.$$

Thanks in advance!

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closed as off-topic by mickep, Did, Hurkyl, user99914, Winther Sep 14 '15 at 21:48

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  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – mickep, Did, Hurkyl, Community, Winther
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  • $\begingroup$ $\lvert z + 1 \rvert = \lvert \overline{z} - 1 \rvert$ gives you $\lvert z + 1 \rvert = \lvert \overline{z} - 1 \rvert$. $\endgroup$ – Daniel Fischer Sep 14 '15 at 18:34
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    $\begingroup$ It is helpful to use $\lvert \overline{w} \rvert = \lvert w\rvert$ for all $w\in \mathbb{C}$ once. $\endgroup$ – Daniel Fischer Sep 14 '15 at 18:37
  • $\begingroup$ Hi and welcome to Math.SE. It is mandatory to give your thoughts on the problem. What have you tried? What is the problem? Have you seen something similar before? $\endgroup$ – mickep Sep 14 '15 at 18:38
  • $\begingroup$ General tip: When presented with a problem like this, think about what you know. A few "facts" that spring to my mind immediately are: $|w|^2=w\bar{w}$, and for $w=x+iy$ then $\bar{w}=x-iy$. See also Daniel Fischer's comment. Apply what you know and see what follows logically. Hopefully it will be the answer! $\endgroup$ – Pixel Sep 14 '15 at 20:27
  • $\begingroup$ Why is this put on hold as off topic? This is hardly an off topic question. $\endgroup$ – Trogdor Sep 15 '15 at 5:03
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let $z=x+yi$
Then we have $$|(x+1)+yi| = |(x-1)-yi| $$ $$(x+1)^2 +y^2= (x-1)^2 +y^2 $$ $$4x=0$$ Then z is set of complex numbers where $$R(z) = 0$$

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  • $\begingroup$ |(x+1)+yi|=|(x−1)−yi| what happens to the i here? (x+1)^2+y^2=(x−1)^2+y^2 $\endgroup$ – Valur Marvin Sep 14 '15 at 19:06
  • $\begingroup$ How do you find norm or length of complex number? $\endgroup$ – Salihcyilmaz Sep 16 '15 at 8:30
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$$(z+1)(z+1)^* = (z^* - 1)(z^* - 1)^*$$ $$|z|+z+z^*+1=|z|-z-z^*+1$$ $$2(z+z^*)=0$$ So basically, $z$ is any purely imaginary number.

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An hint for a geometric view:

using the isomorphism between $\mathbb{C}$ and $\mathbb{R}^2 $ look at $z$ and $\bar z$ as two points symmetric with respect to to the $x-$axis. Your equation state that this two points are equidistant from $(-1,0)$ and $(1,0)$, so they must be on the $y-$axis.

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