1. Interpret the expression

We are given

\qquad \omega=\frac{5+3z}{5(1-z)}\,z.$$ So $$\omega=\frac{z(5+3z)}{5(1-z)}.$$ We need to determine which inequality must be true. --- 2. **Rewrite in a more useful form** Let $$w=\frac{z}{1-z}.$$ Then $$z=\frac{w}{1+w}.$$ But instead of substituting back fully, observe directly: $$\omega=\frac{z(5+3z)}{5(1-z)}=\frac{z}{1-z}\cdot \frac{5+3z}{5}=w\cdot \frac{5+3z}{5}.$$ A cleaner way is to express everything in terms of $w$ using $$z=\frac{w}{1+w}, \qquad 1-z=\frac{1}{1+w}.$$ Then $$\omega=\frac{z(5+3z)}{5(1-z)} =\frac{\frac{w}{1+w}\left(5+3\frac{w}{1+w}\right)}{5\cdot \frac{1}{1+w}}.

This simplifies to

$$=\frac{w(5+8w)}{5(1+w)}.$$

But an even better simplification comes from polynomial division in the original form:

$5\omega=\frac{5z+3z^2}{1-z}.$ Now divide: $\frac{5z+3z^2}{1-z}=-3z-8+\frac{8}{1-z}.$ Hence

$$$$

This is still not ideal.

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3. Use the standard disk-to-half-plane transformation

For $|z|<1$, the quantity $u=\frac{1+z}{1-z}$ satisfies $\Re(u)>0.$ This is a standard result because the map $\dfrac{1+z}{1-z}$ sends the unit disk to the right half-plane.

Now express $\omega$ in terms of $u$.

From $u=\frac{1+z}{1-z},$ we get $z=\frac{u-1}{u+1}.$ Substitute into $\omega=\frac{z(5+3z)}{5(1-z)}.$

First, $1-z=1-\frac{u-1}{u+1}=\frac{2}{u+1}.$ Also,

=\frac{5(u+1)+3(u-1)}{u+1} =\frac{8u+2}{u+1}=\frac{2(4u+1)}{u+1}.$$ Therefore $$\omega=\frac{\frac{u-1}{u+1}\cdot \frac{2(4u+1)}{u+1}}{5\cdot \frac{2}{u+1}} =\frac{(u-1)(4u+1)}{5(u+1)}.$$ Expand numerator: $$ (u-1)(4u+1)=4u^2-3u-1.$$ Now divide by $u+1$: $$4u^2-3u-1=(u+1)(4u-7)+6.$$ So $$\omega=\frac{4u-7}{5}+\frac{6}{5(u+1)}.$$ This is still somewhat messy, but enough for checking real parts. --- 4. **A much simpler substitution** Let $$t=\frac{z}{1-z}.$$ Then since $$z=\frac{t}{1+t},$$ we have $$|z|<1 \iff \left|\frac{t}{1+t}\right|<1.

Squaring modulus: $|t|^2<|1+t|^2=1+|t|^2+2\Re(t),$ which gives $\Re(t)>-\frac12.$

Now express $\omega$ in terms of $t$. Since $z=\frac{t}{1+t},$ we get $5+3z=5+\frac{3t}{1+t}=\frac{5+8t}{1+t}.$ Thus

$$$$

So

$$$$

Now divide:

$$$$

Hence

$$$$

Let $t=x+iy, \qquad x> -\frac12.$ Then

$$\left(\frac{3}{1+t}\right)=\frac{3(1+x)}{(1+x)^2+y^2}.$$

Therefore

$$$$

This expression is not immediately constant, so let us test the options directly by examples and by identifying what is always true.

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5. Test options by choosing valid values of $z$

Since $|z|<1$, any small real $z$ is allowed.

Check option C: $5\Re(\omega)>4$

Take $z=0$. Then $\omega=0,$ so $5\Re(\omega)=0,$ which is not greater than $4$. So C is false.

Check option D: $5\Re(\omega)>1$

Again take $z=0$. Then $5\Re(\omega)=0,$ which is not greater than $1$. So D is false.

Thus the stored answer cannot be correct.

Check option A: $4\Im(\omega)>5$

Take $z=0$ again. Then $\Im(\omega)=0,$ so $4\Im(\omega)=0,$ not greater than $5$. So A is false.

Check option B: $5\Im(\omega)<1$

We must see whether this is always true.

Take a valid point, say $z=0.9i$ (note $|z|=0.9<1$). Then $\omega=\frac{0.9i(5+2.7i)}{5(1-0.9i)}.$ Numerically this gives an imaginary part around $0.69$, so $5\Im(\omega)\approx 3.45>1.$ Hence B is also false.

So none of the four options is universally true.

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6. Conclusion

Using the simplest counterexample $z=0$: $\omega=0.$ Then

• $4\Im(\omega)=0$, so A is false.
• $5\Im(\omega)=0<1$, so B is true for this one value, but not universally.
• $5\Re(\omega)=0$, so C is false.
• $5\Re(\omega)=0$, so D is false.

And since option B also fails for some other valid $z$ (e.g. $z=0.9i$), no option is correct.

Therefore the stored answer $D$ is incorrect.