Question #7290

(5+i3) + (2+i8) (convert to polar form)

Expert's answer

(5+i3) + (2+i8) = (5+2) + (i3+i8) = 7 + i11 = ...

x = 7>0, y = 11>0;

r = sqrt(x²+y²) = sqrt(7²+11²) = sqrt(170)

α = arctan(y/x) = arctan(11/8)

... = r(cosα + isinα) = r(cosα + isinα) = sqrt(170)*(cos(arctan(11/8)) + i*sin(arctan(11/8))).

x = 7>0, y = 11>0;

r = sqrt(x²+y²) = sqrt(7²+11²) = sqrt(170)

α = arctan(y/x) = arctan(11/8)

... = r(cosα + isinα) = r(cosα + isinα) = sqrt(170)*(cos(arctan(11/8)) + i*sin(arctan(11/8))).

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