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

1 Answer
Monzur R.
Jul 15, 2017

(-5-2i) +(7-6i) = 2sqrt17(cos1.33-isin1.33)(−5−2i)+(7−6i)=2√17(cos1.33−isin1.33)

Explanation:

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

This can be written in trigonometric form as:

a+bi = r(cosvartheta + i sinvartheta)a+bi=r(cosϑ+isinϑ) where r=sqrt (a^2+b^2)r=√a2+b2 and vartheta = arctan(y/x)ϑ=arctan(yx)

sqrt(2^2+(-8)^2) = 2sqrt17√22+(−8)2=2√17

arctan(-8/2) = -1.33arctan(−82)=−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)

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