(vi) z4+3z2−4z^4 + 3z^2 - 4z4+3z2−4 Solution Let u=z2:u2+3u−4=(u+4)(u−1)So:z4+3z2−4=(z2+4)(z2−1)Factor each:z2+4=(z−2i)(z+2i)z2−1=(z−1)(z+1)Therefore:z4+3z2−4=(z−1)(z+1)(z−2i)(z+2i)\begin{aligned} & \boxed{\text{Let } u = z^2:} \\ \\ & u^2 + 3u - 4 = (u + 4)(u - 1) \\ \\ & \boxed{\text{So:}} \\ & z^4 + 3z^2 - 4 = (z^2 + 4)(z^2 - 1) \\ \\ & \boxed{\text{Factor each:}} \\ & z^2 + 4 = (z - 2i)(z + 2i) \\ & z^2 - 1 = (z - 1)(z + 1) \\ \\ & \boxed{\text{Therefore:}} \\ & z^4 + 3z^2 - 4 = (z - 1)(z + 1)(z - 2i)(z + 2i) \end{aligned}Let u=z2:u2+3u−4=(u+4)(u−1)So:z4+3z2−4=(z2+4)(z2−1)Factor each:z2+4=(z−2i)(z+2i)z2−1=(z−1)(z+1)Therefore:z4+3z2−4=(z−1)(z+1)(z−2i)(z+2i)