just double checking my answer - Rectangular & Polar form

[Farvahar021]

Hey, its me again , just wanted to double check my answer here to make sure im doing it right.

question is work out the equation 1/Zt = 1/Z1 + 1/Z2 in a) rectangular form and then b) polar form
when Z1 = 1/(15+j) and Z2 = 1/(6-j3)
and below is how i done it

sorry for bad hand writing or if its messy
(by the way 3rd line far right bottom of the fraction its 1521j^2)

 

[mmm4444bot]

1/Zt = 1/Z1 + 1/Z2

when Z1 = 1/(15+j) and Z2 = 1/(6-j^3)

Please note the grouping and caret symbols added in red.

Firstly, some notes about texting notation.

Typing 1/15+j means this:

\(\displaystyle \frac{1}{15} + j\)

If the denominator is 15+j, then you MUST type grouping symbols to indicate that.


The caret symbol ^ (shift 6, on most keyboards) is used to denote exponents.


Secondly, if we substitute 1/(15 + j) for Z1 and 1/(6 - j^3) for Z2 into the expression 1/Z1 + 1/Z2, the resulting expression is:

15 + j + 6 - j^3

(I stopped reading your work, at the first line.)

 

[Farvahar021]

Re: just double checking my answer - Rectangular & Polar for

It wasn't j^3 its just j*3 and yeah sorry forgot about brackets

 

[pka]

Re: just double checking my answer - Rectangular & Polar for

This may far more that you want to know. If so ignore this post.
\(\displaystyle \frac{1}{z} = \frac{{\overline z }}{{\left| z \right|^2 }}\)

Thus \(\displaystyle \frac{1}{{15 + i}} = \frac{{15 - i}}{{126}}\)

Now for polar form.
Suppose that \(\displaystyle z = \left| z \right|\exp \left( {\phi i} \right)\) where \(\displaystyle \phi = \text{Arg}(z)\).
Therefore, \(\displaystyle \frac{1}{z} = \frac{{\exp \left( { - \phi i} \right)}}{{\left| z \right|}}\).