Using ReplaceAll (/.) with complex variables?

Questions
1

I have hopefully a simple question. (I’m using home version of Mathematica 7)

If define a simple complex matrix

test = {{a + Ib, b + Ic}, {d + Ie, e + If}}

and do a simple replacement

test /. {e -> 1, f -> 2}

How come the complex parts using the symbols e and f are not replaced? The above yields

{{a + Ib, b + Ic}, {d + Ie, 1 + If}}

If I do something simple

test2 = a + I b

test2 /. {a -> 4, b -> 2}

Yields: 4 + 2 I

as expected, but if I call up the [[2,2]] element of the array above and try the ReplaceAll it doesn’t work

test6 = test[[2, 2]]

Yields: e + If

but

test6 /. {e -> 3, f -> 5}

Yields: 3 + If

So how come the complex part doesn’t get replaced? How can I do this replacement?

2

Hint:

Ie

is a single symbol.

I*e

or

I e

is the product of two things.

2I

or

2*I

or

2 I

also happen to be the product of two things.

That should be enough.

3

You could’ve noticed it from the coloring already; unknown symbols are blue (or green inside e.g. Module) (a,b,c,d,e,f), known symbols are black (I)…

Note that you can’t use replace(all) on complex numbers:

3 + 4 I /. 4 -> 5

as they are treated as an ‘atom’, so it will just return 3 + 4I.

4

This is not the whole story. ReplaceAll (and related functions) use the FullForm for pattern matching, and to replace a part of the complex number one needs to use this form. It is true that 3 + 4 I is atomic but one may use the head of this expression to get into its parts

FullForm[3 + 4 I]

returns

Complex[3,4]

therefore

3+ 4 I /. Complex[x_,4]:> Complex[x,12]

returns

3 + 12 I

as expected

5

Thanks Bill, for some reason I thought that I was reserved, so Ie and If would be a product to two things.

6

Sorry about the coloring, don’t really know how that happened. a,b,c,d,e,f were suppose to be unknown symbols.

7

Your thinking that I is reserved is correct, but that isn’t sufficient to override the principle that anything beginning with a letter followed by any number of letters and digits will be considered to be one symbol. This kind of problem shows up again and again and again and again…

8

Without any disrespect, whats the fuzz? The OP’s replacements does seem to work (provided there is a space left between I and the variable symbol):

In[1]:= test = {{a + I b, b + I c}, {d + I e, e + I f}}

Out[1]= {{a + I b, b + I c}, {d + I e, e + I f}}

In[2]:= test /. {e -> 1, f -> 2}

Out[2]= {{a + I b, b + I c}, {I + d, 1 + 2 I}}
9

Because you wrote it correctly. Compare the inputs (and the coloring).

{{a + Ib, b + Ic}, {d + Ie, e + If}}
{{a + I b, b + I c}, {d + I e, e + I f}}
10

That is not your fault, that is caused because you need a space between symbols:

If
Ie

versus

I f
I e