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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

It does not display as just i. It is displaying as the esc-ii-esc but the resolution of your screen is blurring the character so you think it is just an i. Use the magnification popup on the lower right hand side of the notebook to increase the magnification to see this.

Read this to understand this better:

http://reference.wolfram.com/language/ref/I.html

POSTED BY: David Reiss
Posted 10 years ago

Thanks to everyone for their answers. I understand, however, why if I write (1+ii) (3-ii) Mathematica replies 4 + 2i (only i)? Only when I change the output Font size appears 4 + 2ii. Should not keep Mathematica the input format (if input I output I, if input ii output ii,...)?

Note: ii = Esc ii Esc

Thanks again.

POSTED BY: rionilo
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

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.

POSTED BY: Marvin Ray Burns

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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