iztrans giving unexpected outputs

My textbook says that the invers Z transform is

My code says the invers is simply k

syms z k
E1 = z/((z-1)^2);
e1_k=iztrans(E1,k)

e1_k = 

k

[Paul]

Assuming the inv(z) to the left of the bracketed expression means "inverse z-transform", I don't see how the expression from the textbook could be correct. Even if there is typo and the denominator should be (z+1)^2, it's still incorrect (missing negative sign).

[Torsten]

Wolfram Alpha gets the same result as MATLAB does:

https://www.wolframalpha.com/input?i=inverse+Z+transform+calculator&assumption=%7B%22F%22%2C+%22InverseZTransformCalculator%22%2C+%22transformfunction%22%7D+-%3E%22z%2F%28-1+%2B+z%29%5E2%22&assumption=%7B%22F%22%2C+%22InverseZTransformCalculator%22%2C+%22variable1%22%7D+-%3E%22z%22&assumption=%7B%22F%22%2C+%22InverseZTransformCalculator%22%2C+%22variable2%22%7D+-%3E%22k%22

[Aidan]

I tested more than just the example given and they were all way off, ;--; I feel like there must be some context for these problems im missing to use this function

[Paul]

Keep in mind that iztrans only applies to inverting the unilateral z-transform. If your book examples are for the bilateral z-transform, then there could be an issue. But that's not the issue for the example cited in the question.

Answer 1

[Sam Chak]

Another way is to provide a proof by contradiction. Misprints can happen due to a lack of proofreading.

syms n

f1  = n*(-1)^n;
ztrans(f1)

ans = 

f2  = n;
ztrans(f2)

ans = 

[Aidan]

I went back and tested more problems in the book, non of the answers match what the function returns ;--;

[Sam Chak]

@Aidan, If it is deemed worthwhile for the MathWorks team to investigate the issue, you may take a screenshot of other z-transform problems from the textbook.