How do you divide #( 6i+4) / (-2i -1 )# in trigonometric form?

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How do you divide #( 6i+4) / (-2i -1 )# in trigonometric form?

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Answer 1 · 2016-01-20T15:52:23

evaluate the following:
#C_3 = sqrt(52)/sqrt(5) angle(26) #
#C_3 = sqrt(52/5)cos26 + i sqrt(52/5)cos26 #

Explanation:

Use the polar form:
#C_1 = 4 + 6i; |C_1| = sqrt(36+16); theta_1 = tan^_1 (6/4) = 89 #
#C_2 = -1 - 2i; |C_2| = sqrt(1+4); theta_2 = tan^_1 (2/1) = -63 #
#C_3 = sqrt(52)/sqrt(5) angle(89-63) #

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How do you divide #( 6i+4) / (-2i -1 )# in trigonometric form?

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