In a normal $2D$ graph the $y$-axis represent the height, $x$-axis the weight. When we want to refer to a certain point on the graph we say $(x, y)$. With complex numbers the case is different, have a look at this graph:


We have now imaginary axis and real axis and that make sense. I however don't understand why points are written like $(1+2i)$? Adding these two points together means that we are converting this $2D$ graph into a $1D$. Why not write them like $(1, 2i)$ instead?

  • 1
    $\begingroup$ Because $1+2i$ is a complex number, $(1,2i)$ is an ordered pair of complex numbers. $\endgroup$ Feb 23, 2018 at 8:16
  • $\begingroup$ @LordSharktheUnknown Thanks $\endgroup$ Feb 23, 2018 at 8:19
  • $\begingroup$ My bet that the OP meant $(1,2)$. $\endgroup$
    – user65203
    Feb 23, 2018 at 9:09

2 Answers 2


What is happen there is an identifiction. To see the complex numbers as points in a plane you identify $\mathbb{C}$ with $\mathbb{R}^2$ by $x+iy \mapsto (x,y)$.


$(a,b)$ and $a+ib$ are strictly equivalent. They express that the complex numbers have two components (the real and imaginary parts). Due to the presence of the $i$ factor, you can always retrieve them from a sum.

The notation $a+ib$, together with the $i^2=-1$ convention is very handy as it allows to use the standard computation rules:

$$(a,b)+(c,d)=(a+c,b+d)\equiv(a+ib)+(c+id)=(a+c)+i(b+d)$$ $$(a,b)\cdot(c,d)=(ac-bd,ad+bc)\equiv(a+ib)\cdot(c+id)=ac+iad+ibc+i^2bd\\ =(ac-bd)+i(ad+bc)$$

You can also consider that you are working with 2D vectors,

$$(a,b)=a\cdot(1,0)+b\cdot(0,1)=a\cdot\mathbb1+b\cdot\mathbb i$$ with $$\mathbb 1\equiv (1,0),\\\mathbb i\equiv(0,1),$$

and a special product operation such that

$$ \begin{array}{c|cc}\times&\mathbb 1&\mathbb i\\\hline\mathbb 1&\mathbb 1&\mathbb i\\\mathbb i&\mathbb i&-\mathbb 1\end{array}\equiv \begin{array}{c|cc}\times&(1,0)&(0,1)\\\hline(1,0)&(1,0)&(0,1)\\(0,1)&(0,1)&(-1,0)\end{array}. $$

This notation is made possible by the fact that $\mathbb C+,\times$ forms a field.


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