Hi.

Just curious though but how would you interpret the polar conversion of

(8+5i)/(3+4i) on the Ti-inspire cas, cause I get this solution of e^(-i〖tan〗^(-1 (17/44))*√89/5,

I understand the e to the power, but I don’t understand the tan part?

For some reason, VCAA uses \(\text{cis}\) notation for polar form, which literally no one beyond VCE uses. Essentially, \[z=r\text{cis}(\theta)=re^{i\theta},\] where \(r=|z|\), and \(\theta=\arg(z)\). Note that the CAS will always give the principal argument of \(z\).

Thus, when your CAS gives \[u=\frac{8+5i}{3+4i}=\frac{\sqrt{89}}{5}e^{-i\arctan(17/44)},\] this means that \[|u|=\frac{\sqrt{89}}{5}\quad \text{and}\quad \text{Arg}(u)=-\arctan\left(\frac{17}{44}\right).\]

Also how do I write equations in these forums, as I'm having to copy equations from word, and their coming across as not so easy to read once I create a post. Thanks

You can copy equations from Microsoft Word. It will produce some code in LaTeX, which is used to display maths on pages.

You'll need to copy your code and put \(\texttt{\[}\) before it, and \(\texttt{\]}\) after it.

For example, the code \[\texttt{\[\int_{-\infty}^\infty e^{-x^2}dx=\sqrt{\pi}\]}\]will produce \[ \int_{-\infty}^\infty e^{-x^2}dx=\sqrt{\pi}\] Alternatively, you can check out

Rui's amazing \(\TeX\) guide to learn how to write the code yourself (which is a lot faster).