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Complex numbers: i vs I

Posted 11 years ago
In[8]:= (1 + I)* (1 - 3 I)      <===== upper case

Out[8]= 4 - 2 i       <=======  lower case

With upper case I in In, it works, but uses the lower case i in Out.

In[11]:= (1 + i)*(1 - 3 i)    <==== lower case

Out[11]= (1 - 3 i) (1 + i)   <===== not multiply?

With lower case i in IN, it doesn't Works.

It seems inconsistent (or not?)

thanks in advance.

POSTED BY: rionilo
5 Replies
Posted 11 years ago

If you look closely at your Out[8] you will see that the lower case i you refer to is different than the usual lower case i. It is Mathematicas output for I. (When the code is pasted in here as a code block, it comes through as I.)

IN with no space between I and N is a symbol. With a space in between, it is I multiplied by N. With 3I or 3N, Mathematica understands it as multiplication, since symbol names cannot begin with a number.

The multiplication example you give can be expanded if you want to see the multiplication carried out. But note that lower case i is not the imaginary unit.

Also note that * is not needed to express multiplication. It is also best to begin all symbol names with lower case, since Mathematica uses upper case for reserved words. For example, N is a built in function for evaluating an argument to a numerical value .

In[1]:= (1 + I) (1 - 3 I)

Out[1]= 4 - 2 I

In[2]:= IN

Out[2]= IN

In[3]:= I N

Out[3]= I N

In[4]:= (1 + i) (1 - 3 i)

Out[4]= (1 - 3 i) (1 + i)

In[5]:= Expand[(1 + i) (1 - 3 i)]

Out[5]= 1 - 2 i - 3 i^2
POSTED BY: David Keith
Posted 11 years ago
POSTED BY: rionilo
POSTED BY: David Reiss

The Out statement uses a special, somewhat italicized, lower case i that you can find in the Basic Math Assistant palette.

The same is true for lower case e.

Lower case i is just the symbol i. Often used as a summation index. Imaginary unit is written capitol I.

POSTED BY: S M Blinder
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