MATHS SOLUTION

Supposez =2+Siandw=5-2i.Then =wz z+w=(2+3i)+(5-2t)=2+5+3i-2t=7+i zw=(2+3i)(5-2i)=10+15t-4i-6t2=16+Hi z=2+Si=2 w 5—2i 2+3t 3i and w=5-2t=5+2i (5-2i)(2-3i)_4-19t_i._31• (2+3i)(2-3t) Justastherealnumberscanberepresentedbythe pointsonaline,thecomplexnumberscanberepresented bythepointsintheplane.Specifically,weletthepoint (a,b)intheplanerepresentthecomplexnumberz=a+bi, i.e.whoserealpartisaandwhoseimaginarypartisb.The absolutevalueofz,written|z|,isdefinedasthedistance fromztotheorigin: \z\ =V^T&^ Notethat|z|isequaltothenormofthevector(a,6).Also,\z Example1.8: Supposez-2+3iandw=12-5i.Then =V4+9=v’iSand|wl 13 13 ZZ. 13^

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