G(jw)=s-0.5/(s+1)

You probably mean G(s) = s - 0.5/(s + 1) and you want G(jω), in which case you just replace s by jω.

 G(jω) = jω - 0.5/(jω + 1)

.

G(jw)=s-0.5/(s+1)

$$\small{\text{$ \begin{array}{rcl} G(j\omega) &=& j\omega-\dfrac{0.5}{j\omega+1}\\\\ G(j\omega) &=& j\omega-\dfrac{0.5}{1+j\omega}\cdot \left( \dfrac{1-j\omega}{1-j\omega} \right) \\\\ G(j\omega) &=& j\omega-\dfrac{0.5(1-j\omega)}{1-j^2\omega^2} \quad | \quad j^2 = -1\\\\ G(j\omega) &=& j\omega-\dfrac{0.5-0.5 j\omega}{1+\omega^2}\\\\ G(j\omega) &=& j\omega+\dfrac{-0.5+0.5 j\omega}{1+\omega^2}\\\\ G(j\omega) &=& j\omega+ \dfrac{0.5 j\omega}{1+\omega^2} -\dfrac{0.5}{1+\omega^2}\\\\ G(j\omega) &=& -\dfrac{0.5}{1+\omega^2} + \left( \omega+ \dfrac{0.5 \omega}{1+\omega^2} \right) j \\\\ \end{array} $}}$$

Thanks Alan and Heureka :)

Heureka has worked out the answer if j is an imaginary number   $$j=\sqrt{-1}$$                

Sometimes j is used instead of i to mean   $$\sqrt{-1}$$

I have a question.

If you get rid of an irrational number in the denominator.  Then you are rationalizing the denominator.

If you get rid of an imaginary number in the denominator does that mean you are realizing the denominator?

It sounds funny but it is a real question :))

$$\\\small{\text{ The product of a complex number and its conjugate is a real number, and is always positive.}}\\\\ $ \small{\text{$ \begin{array}{rcl} (a + bi)(a - bi) &=& a^2 + abi - abi - b^2 \textcolor[rgb]{1,0,0}{i^2} \qquad (i^2=-1)\\ &=& a^2 - b^2 (\textcolor[rgb]{0,0,1}{-1}) ~~$ (the middle terms drop out)$ \\ &=& a^2 + b^2 ~~$ Answer $ \end{array} $}}$$

Thanks Heureka but I was just looking for a word or a term.

Like "rationalizing the domoninator" when dealing with surds.

Maybe there isn't one for complex numbers.  Maybe it just comes under the general heading of simplifying 

Hallo Melody,

to devide a complex Number  z1 / z2 this is the easiest way to do.