How do you add $(8 - 3 i) + (7 + 4 i)$ $(8-3i)+(7+4i)$ in trigonometric form?

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Answer 1 · 2018-05-19T05:29:39

$(\sqrt{226}, tan (\frac{1}{15}))$ $(sqrt226, tan(1/15))$

If you add them together you get:

$15 + i$ $15 + i$

I'm not sure what you mean "trigonometric form" but to convert to polar form $(r, θ)$ $(r, theta)$ you have

Standard complex form: $a x + b i$ $ax +bi$

$r = \sqrt{a^{2} + b^{2}}$ $r = sqrt(a^2+b^2)$

$r = \sqrt{15^{2} + 1^{2}}$ $r = sqrt(15^2+1^2)$

$r = \sqrt{226}$ $r=sqrt226$

$θ = tan (\frac{b}{a})$ $theta = tan(b/a)$

$θ = tan (\frac{1}{15})$ $theta = tan(1/15)$

How do you add (8−3i)+(7+4i)(8-3i)+(7+4i) in trigonometric form?