How do you write the following in trigonometric form and perform the operation given $\frac{5}{2 + 3 i}$ $5/(2+3i)$ ?

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Answer 1 · 2017-06-03T15:17:07

$\frac{5}{2 + 3 i} = \frac{10}{13} - i \frac{15}{13}$ $5/(2+3i)=10/13-i15/13$

Here two numbers are involved

One is $5$ $5$ which is easy to write in trigonometric form as

$5 = 5 (1 + i 0) = 5 ({cos 0}^{\circ} + i {sin 0}^{\circ})$ $5=5(1+i0)=5(cos0^@+isin0^@)$ or $5 e^{i 0}$ $5e^(i0)$

For $2 + 3 i$ $2+3i$ , we know $\sqrt{2^{2} + 3^{2}} = \sqrt{4 + 9} = \sqrt{13}$ $sqrt(2^2+3^2)=sqrt(4+9)=sqrt13$

Hence $2 + 3 i = \sqrt{13} (\frac{2}{\sqrt{13}} + i \frac{3}{\sqrt{13}})$ $2+3i=sqrt13(2/sqrt13+i3/sqrt13)$

Now let $sin α = \frac{3}{\sqrt{13}}$ $sinalpha=3/sqrt13$ and $cos α = \frac{2}{\sqrt{13}}$ $cosalpha=2/sqrt13$ , then

$2 + 3 i = \sqrt{13} (cos α + i sin α) = 5 e^{i α}$ $2+3i=sqrt13(cosalpha+isinalpha)=5e^(ialpha)$

and $\frac{5}{2 + 3 i} = \frac{5 e^{i 0}}{\sqrt{13} e^{i α}}$ $5/(2+3i)=(5e^(i0))/(sqrt13e^(ialpha)$

= $\frac{5}{\sqrt{13}} \times e^{0 - i α} = \frac{5}{\sqrt{13}} e^{- i α}$ $5/sqrt13xxe^(0-ialpha)=5/sqrt13e^(-ialpha)$

= $\frac{5}{\sqrt{13}} (cos (- α) + i sin (- α))$ $5/sqrt13(cos(-alpha)+isin(-alpha))$

= $\frac{5}{\sqrt{13}} (cos α - i sin α)$ $5/sqrt13(cosalpha-isinalpha)$

= $\frac{5}{\sqrt{13}} (\frac{2}{\sqrt{13}} - i \times \frac{3}{\sqrt{13}})$ $5/sqrt13(2/sqrt13-ixx3/sqrt13)$

= $\frac{5}{13} (2 - 3 i) = \frac{10}{13} - i \frac{15}{13}$ $5/13(2-3i)=10/13-i15/13$

How do you write the following in trigonometric form and perform the operation given 5/(2+3i)52+3i5/(2+3i)?