Question #7291

(6+i4) - (3+i8) convert to polar form

Expert's answer

(6+i4) - (3+i8) = (6-3) - (i4+i8) = 3 - i12 = ...

x = 3>0, y = 12>0;

r = sqrt(x²+y²) = sqrt(3²+12²) = sqrt(153)

α = arctan(12/3) = arctan(4)

... = r(cosα + isinα) = r(cosα + isinα) = sqrt(153)*(cos(arctan(4)) + i*sin(arctan(4))).

x = 3>0, y = 12>0;

r = sqrt(x²+y²) = sqrt(3²+12²) = sqrt(153)

α = arctan(12/3) = arctan(4)

... = r(cosα + isinα) = r(cosα + isinα) = sqrt(153)*(cos(arctan(4)) + i*sin(arctan(4))).

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