I need help expressing in (a+ib)form & simplifying pleas

[nil101]

\(\displaystyle \L


\begin{array}{l}
{\rm If }z = R + \iota \omega L - \frac{1}{{\iota \omega C}} \\
{\rm express }z{\rm in (}a + \iota b){\rm form when }R = 34,L = 6,C = 0.05 {\rm and } \omega = 5, \\
{\rm hence simplify }\frac{{{\rm \bar z}}}{{{\rm z}^{\rm 2} }} \\
\end{array}\)

 

[Unco]

express z in (a + ib)

Substitute in the values and add like terms to simplify. Note that \(\displaystyle \L \mbox{ \frac{ 1 }{i} = \frac{ i }{i^2} = -i}\).

hence simplify \(\displaystyle \L \frac{ \bar z}{z^2}\)

Factor out 34 in the numerator and 34*34 in the denominator. Expand the denominator.

 

[nil101]

Please check my solution

Please can you check this?

\(\displaystyle \L

z = 34 + 30\iota - \frac{1}{{0.25\iota }} = 34 + 30i - (4)( - \iota ) = 34 + 34\iota\)



\(\displaystyle \L

\bar z = 34 - 34\iota = 34\left( {1 - \iota } \right)\)


\(\displaystyle \L

z^2 = \left( {34 + 34i} \right)^2 = 34^2 \left( {1 + \iota } \right)^2\)


\(\displaystyle \L

\frac{{\bar z}}{{z^2 }} = \frac{{34\left( {1 - \iota } \right)}}{{34^2 \left( {1 + \iota } \right)^2 }} = \frac{{\left( {1 - \iota } \right)}}{{34\left( {1 + \iota } \right)^2 }} = \frac{{\left( {1 - \iota } \right)}}{{34\left( {1 + 2\iota - 1} \right)}} = \frac{{\left( {1 - \iota } \right)}}{{34\left( {2\iota } \right)}} = \frac{{\left( {1 - \iota } \right)}}{{68\iota }}\)

Is this correct?

Thanks

 

[Unco]

Yes, that is absolutely correct!

You can finish it off by putting \(\displaystyle \L \mbox{\frac{\bar z}{z^2}}\) in the form \(\displaystyle \mbox{a + bi}\).