Russell Gillen
2022-01-03
Answered

Since $5i=5{e}^{i\frac{\pi}{2}}$ And $2+i=\sqrt{5}{e}^{i{x}_{1}}$ where $x}_{1}=\mathrm{arctan}\frac{12}{$ , we have $\frac{5i}{2+i}=\sqrt{5}[\mathrm{cos}(\frac{\pi}{2}-{x}_{1})+i\mathrm{sin}(\frac{\pi}{2}-{x}_{1})]$ Now from here how do I continue?

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Vasquez

Answered 2022-01-08
Author has **460** answers

We already have

Continue with

Note: polar form represented by exponential form for simplicity in (1)(2). There is no difference between two forms since r and

user_27qwe

Answered 2022-01-08
Author has **230** answers

Hint

So now let's develop the hint from where you left it i.e

So

And so

nick1337

Answered 2022-01-08
Author has **575** answers

Using polar form (with exponential notation for convenience),

Sketch the

nick1337

Answered 2022-01-08
Author has **575** answers

discarded because this solution will make zero the denominator of the original equation

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