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

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Answer 1 · 2017-06-03T15:17:07

$5/(2+3i)=10/13-i15/13$

Here two numbers are involved

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

$5=5(1+i0)=5(cos0^@+isin0^@)$ or $5e^(i0)$

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

Hence $2+3i=sqrt13(2/sqrt13+i3/sqrt13)$

Now let $sinalpha=3/sqrt13$ and $cosalpha=2/sqrt13$ , then

$2+3i=sqrt13(cosalpha+isinalpha)=5e^(ialpha)$

and $5/(2+3i)=(5e^(i0))/(sqrt13e^(ialpha)$

= $5/sqrt13xxe^(0-ialpha)=5/sqrt13e^(-ialpha)$

= $5/sqrt13(cos(-alpha)+isin(-alpha))$

= $5/sqrt13(cosalpha-isinalpha)$

= $5/sqrt13(2/sqrt13-ixx3/sqrt13)$

= $5/13(2-3i)=10/13-i15/13$

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