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Convert z =cos+i sin into polar form. |
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Answer» let i-1/ cos(180/3)+isin(180/3)= i-1/ cos60+ isin60=i-1/1/2+V3/2i= i-1/1+V3i/2=2(i-1)/1+V3i; rationalizing =2(i-1)/1+V3i x 1- V3i/1-V3i =2[i(i-V3i)-1(1-V3i)/(1-V3i)(1-V3i) using (a-b)(a+b)=a^2-b^2 =2(-1+u+V3i-V3i(-1)/1-3(-1)= 2(-1+i + V3i+V3/1+3= 2((-1+V3)+(i+V3i))/4 =(-1+V3)+i(1+V3)/2= V3-1/2 + I v3+1/2; z=V3-1/2 + I V3+1/2 |
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