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

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

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Answer 1 · 2016-05-18T09:32:52

$(2 + 8 i) + (3 + 4 i) = \sqrt{68} [cos α + i sin α] + 5 [cos β + i s sin β]$ $(2+8i)+(3+4i)=sqrt68[cosalpha+isinalpha]+5[cosbeta+issinbeta]$ ,

where $α = arctan 4$ $alpha=arctan4$ and $β = arctan (\frac{4}{3})$ $beta=arctan(4/3)$ .

Explanation:

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

$a + i b$ $a+ib$ can be written in trigonometric form $r cos θ a + i r sin θ = r (cos θ a + i sin θ)$ $rcosthetaa+irsintheta=r(costhetaa+isintheta)$ ,
where $r = \sqrt{a^{2} + b^{2}}$ $r=sqrt(a^2+b^2)$ and $tan θ = \frac{b}{a}$ $tantheta=b/a$ or $θ = arctan (\frac{b}{a})$ $theta=arctan(b/a)$

Hence $(2 + 8 i) = \sqrt{2^{2} + 8^{2}} [cos α + i sin α]$ $(2+8i)=sqrt(2^2+8^2)[cosalpha+isinalpha]$ or

$\sqrt{68} [cos α + i sin α]$ $sqrt68[cosalpha+isinalpha]$ , where $α = arctan 4$ $alpha=arctan4$ and

$(3 + 4 i) = \sqrt{3^{2} + 4^{2}} [cos β + i sin β]$ $(3+4i)=sqrt(3^2+4^2)[cosbeta+isinbeta]$ or

$5 [cos β + i s sin β]$ $5[cosbeta+issinbeta]$ , where $β = arctan (\frac{4}{3})$ $beta=arctan(4/3)$

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

$\sqrt{68} [cos α + i sin α] + 5 [cos β + i s sin β]$ $sqrt68[cosalpha+isinalpha]+5[cosbeta+issinbeta]$ ,

where $α = arctan 4$ $alpha=arctan4$ and $β = arctan (\frac{4}{3})$ $beta=arctan(4/3)$ .

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

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