To evaluate the definite integral \(\int^{z}_{0}(sin x - cos x) dx\) where \(z = \frac{\pi}{4}\), we can first find the antiderivative of the integrand:

\(\int (sin x - cos x) dx = -cos x - sin x + C\)

where C is the constant of integration. Now, we can use the Fundamental Theorem of Calculus to evaluate the definite integral:

\(\int^{z}_{0}(sin x - cos x) dx = [-cos x - sin x]^{z}_{0}\)

Substituting the values of the limits of integration, we get:

\(\int^{\frac{\pi}{4}}_{0}(sin x - cos x) dx = [-cos x - sin x]^{\frac{\pi}{4}}_{0}\)

Evaluating the expression at the upper and lower limits, we get:

\(\int^{\frac{\pi}{4}}_{0}(sin x - cos x) dx = [-cos(\frac{\pi}{4}) - sin(\frac{\pi}{4})] - [-cos(0) - sin(0)]\)

Simplifying this expression, we get:

\(\int^{\frac{\pi}{4}}_{0}(sin x - cos x) dx = [-\frac{\sqrt{2}}{2} - \frac{\sqrt{2}}{2}] - [-1]\)

\(\int^{\frac{\pi}{4}}_{0}(sin x - cos x) dx = -\sqrt{2} + 1\)

So, the value of the definite integral \(\int^{\frac{\pi}{4}}_{0}(sin x - cos x) dx\) is -\sqrt{2} + 1.