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I have proven that $expr=(1 - i)(\dfrac{1+z}{1-z})^{1/4}\dfrac{\sqrt{ 1-z^2+i z\sqrt{1 - z^2}}}{1+z-i\sqrt{1 - z^2}}=1$ when it is assumed that $z\in R$ and $|z|<1$ and when I also plot the real and imaginary parts of it in that domain, it is indeed equal to $1$, but when I try to use
FullSimplify[expr, Assumptions$\mapsto${$z\in R$ && $Abs[z]<1$}]
in order to try to get $1$, it gives back the same form of the expression.

So, what is the problem and how can I fix this?

EDIT: In Mathematica, I wrote this as
expr = (1 - I) ((1 + z)/(1 - z))^(1/4) Sqrt[ 1 - z^2 + IzSqrt[1 - z^2]]/(1 + z - I*Sqrt[1 - z^2]);
And after getting the result, I used FullSimplify[%,Assumptions -> {z $\in$ R && Abs[z]<1}]

I have proven that $expr=(1 - i)(\dfrac{1+z}{1-z})^{1/4}\dfrac{\sqrt{ 1-z^2+i z\sqrt{1 - z^2}}}{1+z-i\sqrt{1 - z^2}}=1$ when it is assumed that $z\in R$ and $|z|<1$ and when I also plot the real and imaginary parts of it in that domain, it is indeed equal to $1$, but when I try to use
FullSimplify[expr, Assumptions -> {z \[Element] Reals && Abs[z] < 1}]
in order to try to get $1$, it gives back the same form of the expression.

So, what is the problem and how can I fix this?

EDIT: In Mathematica, I wrote this as

expr = (1 - I) ((1 + z)/(1 - z))^(1/4) Sqrt[
     1 - z^2 + I*z*Sqrt[1 - z^2]]/(1 + z - I*Sqrt[1 - z^2]);

And after getting the result, I used FullSimplify[%, Assumptions -> {z \[Element] Reals && Abs[z] < 1}]

I have proven that $expr=(1 - i)(\dfrac{1+z}{1-z})^{1/4}\dfrac{\sqrt{ 1-z^2+i z\sqrt{1 - z^2}}}{1+z-i\sqrt{1 - z^2}}=1$ when it is assumed that $z\in R$ and $|z|<1$ and when I also plot the real and imaginary parts of it in that domain, it is indeed equal to $1$, but when I try to use
FullSimplify[expr, Assumptions$\mapsto${$z\in R$ && $Abs[z]<1$}]
in order to try to get $1$, it gives back the same form of the expression.

So, what is the problem and how can I fix this?

I have proven that $expr=(1 - i)(\dfrac{1+z}{1-z})^{1/4}\dfrac{\sqrt{ 1-z^2+i z\sqrt{1 - z^2}}}{1+z-i\sqrt{1 - z^2}}=1$ when it is assumed that $z\in R$ and $|z|<1$ and when I also plot the real and imaginary parts of it in that domain, it is indeed equal to $1$, but when I try to use
FullSimplify[expr, Assumptions$\mapsto${$z\in R$ && $Abs[z]<1$}]
in order to try to get $1$, it gives back the same form of the expression.

So, what is the problem and how can I fix this?

EDIT: In Mathematica, I wrote this as
expr = (1 - I) ((1 + z)/(1 - z))^(1/4) Sqrt[ 1 - z^2 + IzSqrt[1 - z^2]]/(1 + z - I*Sqrt[1 - z^2]);
And after getting the result, I used FullSimplify[%,Assumptions -> {z $\in$ R && Abs[z]<1}]

I have proven that $expr=(1 - i)(\dfrac{1+z}{1-z})^{1/4}\dfrac{\sqrt{ 1-z^2+i z\sqrt{1 - z^2}}}{1+z-i\sqrt{1 - z^2}}=1$ when it is assumed that $z\in R$ and $|z|<1$ and when I also plot the real and imaginary parts of it in that domain, it is indeed equal to $1$, but when I try to use
FullSimplify[expr, Assumptions$\mapsto${$z\in R$ && $Abs[z]<1$}]
in order to try to get $1$ (because I need to do such simplifications for a bigger calculation I need to do), it gives the same form of the expression back.

So, what is the problem and how can I fix this?

I have proven that $expr=(1 - i)(\dfrac{1+z}{1-z})^{1/4}\dfrac{\sqrt{ 1-z^2+i z\sqrt{1 - z^2}}}{1+z-i\sqrt{1 - z^2}}=1$ when it is assumed that $z\in R$ and $|z|<1$ and when I also plot the real and imaginary parts of it in that domain, it is indeed equal to $1$, but when I try to use
FullSimplify[expr, Assumptions$\mapsto${$z\in R$ && $Abs[z]<1$}]
in order to try to get $1$, it gives back the same form of the expression.

So, what is the problem and how can I fix this?