How do you add $(2+8i)+(3+4i)$ in trigonometric form?

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How do you add $(2+8i)+(3+4i)$ in trigonometric form?

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Answer 1 · 2016-05-18T09:32:52

$(2+8i)+(3+4i)=sqrt68[cosalpha+isinalpha]+5[cosbeta+issinbeta]$ ,

where $alpha=arctan4$ and $beta=arctan(4/3)$ .

Explanation:

Let us first write $(2+8i)$ and $(3+4i)$ in trigonometric form.

$a+ib$ can be written in trigonometric form $rcosthetaa+irsintheta=r(costhetaa+isintheta)$ ,
where $r=sqrt(a^2+b^2)$ and $tantheta=b/a$ or $theta=arctan(b/a)$

Hence $(2+8i)=sqrt(2^2+8^2)[cosalpha+isinalpha]$ or

$sqrt68[cosalpha+isinalpha]$ , where $alpha=arctan4$ and

$(3+4i)=sqrt(3^2+4^2)[cosbeta+isinbeta]$ or

$5[cosbeta+issinbeta]$ , where $beta=arctan(4/3)$

Hence $(2+8i)+(3+4i)$ =

$sqrt68[cosalpha+isinalpha]+5[cosbeta+issinbeta]$ ,

where $alpha=arctan4$ and $beta=arctan(4/3)$ .

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How do you add $(2+8i)+(3+4i)$ in trigonometric form?

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