What is the trigonometric form of (12-2i) (12−2i)?

1 Answer
Jim G.
Mar 11, 2016

2sqrt37 [ cos(-0.165) + isin(-0.165) ]2√37[cos(−0.165)+isin(−0.165)]

Explanation:

Using the following formulae :

• r^2 = x^2 + y^2∙r2=x2+y2

• theta = tan^-1 (y/x)∙θ=tan−1(yx)

here x = 12 and y = - 2

hence r^2 = 12^2 + (-2)^2 = 148 rArr r = sqrt148 = 2sqrt37r2=122+(−2)2=148⇒r=√148=2√37

and theta = tan^-1(-2/12) ≈ -0.165" radians " θ=tan−1(−212)≈−0.165 radians

rArr (12-2i) = 2sqrt37 [cos(-0.165) + isin(-0.165) ]⇒(12−2i)=2√37[cos(−0.165)+isin(−0.165)]

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