How do you divide # (-1+5i)/(3-i) # in trigonometric form?

1 Answer
Shaikh A.
Dec 4, 2016

In Trigonometric form
#sqrt13/sqrt5 [cos (-60.26) +i sin (97.12)]#

Explanation:

In trigonometric form consider Z1 = (1+5i) and Z2=(3-i) and A1 = inv tan (- 5/1) = -78.69 and A2 = inv tan (- 1/3) = -18.43

#mod z1 = sqrt26 and mod z2 = sqrt10#

the number would be
z1/z2 [cos (A1 -A2) + i sin (A1+A2)]

# (sqrt13sqrt2)/(sqrt5sqrt2) [cos (-60.26) + i sin (9.12)] #
# (sqrt13)/(sqrt5) [cos (-60.26) + i sin (9.12)] #

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