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I plugged into google calculator (because that is what I had on hand) $\mathrm{abs}(i)$ and it was said that equaled $1$. Is that true? And if so can I have a proof that it is true?

Note:

As a HS student taking AP Calculus my math proofs are not very advanced but I do understand the non textbook proof (mathematical proof) so if you would answer with a less formal proof I would very much appreciate it.

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  • $\begingroup$ I am sorry if this is off-topic or if I am doing this wrong. It is my first question on this SE and I don't know what is good/bad to ask here $\endgroup$ – Christopher Aug 30 '17 at 1:58
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    $\begingroup$ Look up how modulus of a complex number is defined. $\endgroup$ – Kenny Lau Aug 30 '17 at 1:59
  • $\begingroup$ @ThomasAndrews I fixed that. It was $=1$ $\endgroup$ – Christopher Aug 30 '17 at 2:00
  • $\begingroup$ less formal proof The absolute value $\,|\cdot|\,$ is a positive real number, and the function is multiplicative $\,|ab|=|a|\,|b|\,$. But $|i^2|=|-1|=1\,$, so $\,|i|\,$ is the (real) square root of that i.e. $\,1\,$. $\endgroup$ – dxiv Aug 30 '17 at 2:10
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For a complex number $z$, you can think of $|z|$ referring to the distance between $(0,0)$ and $z$ on a complex plane. (you can think of absolute value as the distance between the origin and your number. e.g. $-3$ and $3$ are both $3$ away from 0)

We find the distance using Pythagorean theorem:

$$ | a + bi | = \sqrt{a^2+b^2} $$

so if we add our values:

$$ \begin{align} &\sqrt{0^2 + 1^2} \\ =&\sqrt{1} \\ =&1 \end{align} $$

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  • $\begingroup$ Shouldn't that be $|a+bi|$? $\endgroup$ – Thomas Andrews Aug 30 '17 at 2:04
  • $\begingroup$ Welcome back to math.SE! $\endgroup$ – Kenny Lau Aug 30 '17 at 2:07
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The absolute value of a complex number $a+bi$ (with $a$ and $b$ real) is defined as $\sqrt{a^2+b^2}$. In the case of $i=0+1i$, we get $|i|=\sqrt{0^2+1^2}=\sqrt{1}=1$.

One motivation for this definition is that if you think of $a+bi$ as representing the point $(a,b)$ in the plane, then $\sqrt{a^2+b^2}$ is the distance from this point to the origin $(0,0)$ (which represents the complex number $0+0i=0$). So the absolute value of a complex number measures "how far it is from $0$", just like the absolute value of a real number.

For some discussion of further motivation behind this definition, you may be interested in this answer of mine.

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This is simply the definition of $|z|$ for complex $z$. If $z=a+bi$ with $a,b$ real, then $$|z|=\sqrt{a^2+b^2}.$$ This is the distance of $z$ to $0$ using the standard Euclidean metric.

When $b=0$, you get the standard real absolute value:

$$\sqrt{a^2}=|a|.$$

One nice feature if this definition is that if $w,z$ are complex numbers, then:

$$|wz|=|w|\cdot|z|.$$

This is not 100% obvious from the definition, and there are reasons related to the "geometry" of complex multiplication that explain this.

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Every complex number $z$ can be uniquely written as $z=re^{i\theta}$ for some $r\in\mathbb R, r\geq 0$ and $\theta\in[0,2\pi)$. The norm of $z$ is defined as $|z|=r$.

Now because $i=1\cdot e^{i\pi}$, we have that $|i|=1$.

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The modulus of a complex number is the square root of the product with its conjugate. This gives: $$\lvert i\rvert=\sqrt{i(-i)}=\sqrt 1.$$

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absolute value/ complex modulus is related to the distance you have to travel between the point and the origin.

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