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InterviewSolution
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Express `sin""pi/5+i(1-cos"" pi/5)`in polar form. |
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Answer» Let `z = "sin"(pi)/(5)+ i(1-"cos"(pi)/(5))`. Let its polar form be `z = r(cos theta + i sin theta)`. Now, `r^(2)=|z|^(2)="sin"^(2)(pi)/(5)+(1-"cos"(pi)/(5))^(2)=("sin"^(2)(pi)/(5)+"cos"^(2)(pi)/(5))+1 -2 "cos"(pi)/(5)` `rArr" "r^(2)=2(1-"cos"(pi)/(5))=4"sin"^(2)(pi)/(10) rArr r=2"sin(pi)/(10)`. Let `alpha` be the acute angle, given by `tan alpha=|(Im(z))/(Re(z))|=|(1-"cos"(pi)/(5))/("sin"(pi)/(5))|=(2"sin"^(2)(pi)/(10))/(2"sin"(pi)/(10)."cos"(pi)/(10))= "tan"(pi)/(10)rArr alpha = (pi)/(10)`. Clearly, the point representing z lies in the first quadrant as `x gt 0 and y gt 0`. `therefore" "arg(z)= theta = alpha = (pi)/(10)`. Thus, `r=2"sin"(pi)/(10)and theta=(pi)/(10)`. Hence, the required polar form is `2"sin"(pi)/(10)("cos"(pi)/(10)+"i sin"(pi)/(10))`. |
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