# precalculus

put this in trigonometric form:

-3-i

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1. Try writing it as the product of a scalar sqrt(10) and an expression of the form e^(i*theta), where theta is an angle in the third quadrant. It will be 180 + arctan 1/3 = 198.435 degrees or 3.4533 radians

sqrt(10)*exp(3.4533 i) = -3 - i

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2. -3 + e[3/2)i pi]

or sqrt 10* exp (i A)

where A = pi + arctan (1/4)= 3.38657 radians

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3. Or maybe they would like with sin A and cos A
where R e^iA = R cos A + i R sin A
then
-3 = R cos A
-1 = R sin A
so
R^2 sin^2 A + R^2 cos^2 A = 1+9 = 10
R^2 (1) = 10
R = sqrt 10 like he said
then cos A = -3/sqrt 10
and sin A = -1/sqrt 10
so
A is in quadrant 3 and is 198.43 degrees or 3.46 radians
so
-3 -i = sqrt 10 * [cos 3.46 + i sin 3.46]

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4. Damian, I think that is what they wanted.
Here is a the same kind of question from earlier.

http://www.jiskha.com/display.cgi?id=1211758996

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