$$z=\sqrt{3+4j}\\ ⇒z^2=3-4j\\ ⇒|z_1|=\sqrt{3^2+4^2}=5\\ tanφ=\frac{Imaginärteil}{Realteil}=\frac{4}{3}\\φ=53,130 \space \text{(1. Quadrant; n=2)}\\ \boxed{allg: z_k=|z_1|^{\frac{1}{n}}\cdot e^{j\frac{φ+2kπ}{n}}(k=0,1)}\\ a)\\ z_0=5^\frac{1}{2}\cdot e^{j\frac{0+53,130}{2}}=\sqrt{5}\cdot e^{j26,565°}\\ ⇒ Polardarstellung\\ z_1=\sqrt{5}(cos(26,565°)+j\space sin(26,565°))\\ z_1=2+5⇒allg. Form\\ b)\\ z_1=\sqrt{5}\cdot e^{j\frac{53,13°+2π}{2}}\\ z_1=\sqrt{5}e^{j206,565°}\\ ⇒z_1=\sqrt{5}(cos(206,565°)+j\space sin(206,565°)){}\\ Z-2=-2-j$$