## Calculus 8th Edition

a) $\overline{z+w}=\overline{z}+\overline{w}$ b) $\overline{zw}=\overline{z} \cdot \overline{w}$ c) $\overline{z^n}=\overline{z}^{n}$
a) Suppose $z=a+bi; w=c+di$ $\overline{z+w}=\overline{(a+bi)+(c+di)}$ $\implies \overline{z+w}=(a+c)-(b+d)i$\ This implies that $\overline{z+w}=\overline{z}+\overline{w}$ b) Here, w ehave $\overline{zw}=\overline{(a+bi)(c+di)}$ $\implies \overline{zw}=ac-bd-bci-adi$ $\implies \overline{zw}=(a-bi)(c-di)$ Thus, we have $\overline{zw}=\overline{z} \cdot \overline{w}$ c) Need to apply De Movire's Theorem such as: $\overline{z^n}=\overline{r^n(\cos n \theta+i \sin n \theta)}$ This implies that $\overline{z^n}=[r(\cos \theta-i \sin \theta)]^n$ Thus, we have $\overline{z^n}=\overline{z}^{n}$