How do you add #(-5-2i)+(7-6i)# in trigonometric form?

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Answer 1 · 2017-07-15T17:31:01

#(-5-2i) +(7-6i) = 2sqrt17(cos1.33-isin1.33)#

#(-5-2i)+(7-6i) = 2-8i#

This can be written in trigonometric form as:

#a+bi = r(cosvartheta + i sinvartheta)# where #r=sqrt (a^2+b^2)# and #vartheta = arctan(y/x)#

#sqrt(2^2+(-8)^2) = 2sqrt17#

#arctan(-8/2) = -1.33#

#therefore 2-8i = 2sqrt17(cos(-1.33) + isin(-1.33))#

Using the properties of the #sin# and #cos# functions, this can be rewritten as:

#2-8i=2sqrt17(cos1.33-isin1.33)#