A complex number is defined by \( z=1+i \). (a) Express \( z \) in polar form. (b) Use de Moivre's theorem to evaluate \( z^{8} \).

To express the complex number \( z = 1+i \) in polar form, we first need to calculate its modulus and argument. The modulus \( r \) is given by \( r = \sqrt{x^2 + y^2} \) where \( x \) and \( y \) are the real and imaginary parts, respectively. Here, \( x = 1 \) and \( y = 1 \), so: \[ r = \sqrt{1^2 + 1^2} = \sqrt{2}. \] Next, we find the argument \( \theta \). Since both \( x \) and \( y \) are positive, \( \theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}(1) = \frac{\pi}{4} \). Therefore, the polar form of \( z \) is: \[ z = r(\cos \theta + i \sin \theta) = \sqrt{2} \left(\cos\frac{\pi}{4} + i \sin\frac{\pi}{4}\right). \] Next, to evaluate \( z^{8} \) using de Moivre's theorem, we have: \[ z^n = r^n \left( \cos(n\theta) + i \sin(n\theta) \right). \] Here, \( n = 8 \), so we compute: \[ r^8 = (\sqrt{2})^{8} = 2^{4} = 16, \] and \[ 8\theta = 8 \cdot \frac{\pi}{4} = 2\pi. \] Thus, \[ z^{8} = 16 \left( \cos(2\pi) + i \sin(2\pi) \right). \] Since \( \cos(2\pi) = 1 \) and \( \sin(2\pi) = 0 \), we have: \[ z^{8} = 16(1 + 0i) = 16. \] In conclusion, \( z \) in polar form is \( \sqrt{2} \left(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}\right) \) and \( z^{8} = 16 \).