\(\eqalign{ & a)\,\overline z = r\left( {\cos \varphi - i\sin \varphi } \right) = r\left( {\cos \left( { - \varphi } \right) + i\sin \left( { - \varphi } \right)} \right) \cr & - z = - r\left( {\cos \varphi + i\sin \varphi } \right) = r\left( {\cos \left( {\pi + \varphi } \right) + i\sin \left( {\pi + \varphi } \right)} \right) \cr & {1 \over z} = {z \over {\overline z .z}} = {1 \over r}\left( {\cos \varphi + i\sin \varphi } \right) \cr & k.z = kr\left( {\cos \varphi + i\sin \varphi } \right)\,\,\text{nếu}\,k > 0 \cr & kz = - kr\left( {\cos \left( {\pi + \varphi } \right) + i\sin \left( {\pi + \varphi } \right)} \right)\,\,\text{nếu}\,\,k < 0 \cr} \)
\(b)\,z = 1 + \sqrt 3 i = 2\left( {{1 \over 2} + {{\sqrt 3 } \over 2}i} \right) = 2\left( {\cos {\pi \over 3} + i\sin {\pi \over 3}} \right)\)
Áp dụng câu a) ta có: \(\overline z = 2\left( {\cos \left( { - {\pi \over 3}} \right) + i\sin \left( { - {\pi \over 3}} \right)} \right)\)
\( - z = 2\left( {\cos {{4\pi } \over 3} + i\sin {{4\pi } \over 3}} \right);\,{1 \over {\overline z }} = {1 \over 2}\left( {\cos {\pi \over 3} + i\sin {\pi \over 3}} \right)\)
\(\eqalign{ & kz = 2k\left( {\cos {\pi \over 3} + i\sin {\pi \over 3}} \right)\,\,\text{nếu}\,\,k > 0 \cr & kz = - 2k\left( {\cos {{4\pi } \over 3} + i\sin {{4\pi } \over 3}} \right)\,\text{nếu}\,\,k < 0 \cr} \)
