Is it possible in various calculations related to the complex plane which also include analytic geometry , calculating distances etc, to omit $i$ and treat the imaginary axis as simply the cartesian $y$ axis with purely real values like we do regularly with all Real $x,y$ axis?
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- 4$\begingroup$ That depends essentially what exactly your "various calculations" are. $\endgroup$hmakholm left over Monica– hmakholm left over Monica2014-07-10 15:35:39 +00:00Commented Jul 10, 2014 at 15:35
- $\begingroup$ @Henning those were stated in my question $\endgroup$Bak1139– Bak11392014-07-10 15:47:39 +00:00Commented Jul 10, 2014 at 15:47
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2 Answers
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The short answer is YES. In topology, $\mathbb{C}$ is equivalent to $\mathbb{R}^2$, so everything you do in $\mathbb{C}$, you can also do it in $\mathbb{R}^2$ by equivalent way.
However, there are nice field properties on $\mathbb{C}$, which means you can treat $(a,b)$ as a single number $a+ib$ to do calculation.
In another word, similar to $\mathbb{R}$, $\mathbb{C}$ is a field, but $\mathbb{R}^2$ is a vector space in general. You can transplant operations on $\mathbb{R}$ to $\mathbb{C}$ easily, while opeartions on vector field are limited.
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3 You can treat the complex numbers as the matrices of the form $$\begin{pmatrix}x&-y\\y&x\end{pmatrix}$$
Then complex multiplication and addition corresponds to the related matrix multiplication and addition.
It's not clear what you mean by "treat the imaginary axis as simply the cartesian y axis" when talking about computations. You can certainly define the complex numbers as the ring $(\mathbb R^2, +,\times)$ with the operations defined as $$(x_1,y_1)+(x_2,y_2)=(x_1+x_2,y_1+y_2)$$ and $$(x_1,y_1)\times(x_2,y_2)=(x_1x_2-y_1y_2, x_1y_2+y_1x_2)$$ That is one way that mathematicians tend to rigorously define complex numbers, in fact...
It's a bit of a misnomer to say you are working "without $i$" here. You have just change notation, but you've really just hidden $i$.
- $\begingroup$ Your minus sign is in the wrong place in your matrix. Your description makes $i\cdot i=1$ instead of $-1$. $\endgroup$Ian– Ian2014-07-10 15:55:11 +00:00Commented Jul 10, 2014 at 15:55
- $\begingroup$ I do that basically every time ;) $\endgroup$Emily– Emily2014-07-10 16:11:59 +00:00Commented Jul 10, 2014 at 16:11
- $\begingroup$ I've reverse the signs to the "usual" representation of the complex numbers as matrices, but actually, it doesn't matter which $y$ gets the minus sign - it is the nature of the conjugation automorphism. It's only if you consider the matrix and $(x,y)^T$ as related that the sign matters, and you want the matrix to act on the left as multiplication (rather than, say, acting on the right on $(x,y)$.) @Ian $\endgroup$Thomas Andrews– Thomas Andrews2014-07-10 16:29:18 +00:00Commented Jul 10, 2014 at 16:29