# Trigonometryfinding the impedance in rectangular and polar form

#### MFletch

##### New member
I dont fully understand how to work out the impedance from the given equation (5j-5)x(11j-11)/(5j-5)+(11j-11). Any help would be greatly appreciated. Thanks.

The answer needs to be in rectangular and polar form.

#### DavidCampen

##### Member
The numerator factors into 55$$(j-1)^2$$ and the denominator to 16(j-1); from here it should be simple to calculate the rectangular form.

#### MFletch

##### New member
Thank you, however, How would that convert to polar and Rec. form?

#### DavidCampen

##### Member
$$\frac{55{(j-1)}^{2}}{16(j-1)} = ?$$

Hint, cancel like terms and you will have the rectangular form.

so the rectangular form will be

$$\frac{55}{16}(j-1)$$

Last edited:

#### MFletch

##### New member
great thank you I've got it!

What would be the best way to find the admittance of Z = (1/5j-1) + (1/2j+6) + (1/4j) in rectangular and polar form?

#### DavidCampen

##### Member
great thank you I've got it!

What would be the best way to find the admittance of Z = (1/5j-1) + (1/2j+6) + (1/4j) in rectangular and polar form?
Add the imaginary parts together and separately add the real parts together to get the simplified rectangular form (aj + b) then you can convert to polar form by finding the magnitude = $\sqrt{{a}^{2}+{b}^{2}}$ and the angle = ${tan}^{-1}(\frac{a}{b})$

#### MFletch

##### New member
I’m not sure I follow

#### DavidCampen

##### Member
"What would be the best way to find the admittance of Z = (1/5j-1) + (1/2j+6) + (1/4j) in rectangular and polar form?"

Add together the 3 three terms given above. The first two terms are$(\frac{1}{5}j-1)$ and $(\frac{1}{2}j+6)$, what is the sum?

Hint:
What is the sum of $\frac{1}{5}j$ and $\frac{1}{2}j$?
What is the sum of -1 and 6?

I see also that this question is asking for admittance which is the reciprocal of impedance. So proceed as above to find the impedance in rectangular form and then find the reciprocal of that.

Last edited:

#### skeeter

##### Well-known member
MHB Math Helper
What would be the best way to find the admittance of Z = (1/5j-1) + (1/2j+6) + (1/4j) in rectangular and polar form?
just checking. do you mean ...

$Z = \left(\dfrac{1}{5} j - 1 \right) + \left(\dfrac{1}{2} j + 6 \right) + \left(\dfrac{1}{4} j \right)$

or ...

$Z = \dfrac{1}{5j - 1} + \dfrac{1}{2j + 6} + \dfrac{1}{4j}$

#### MFletch

##### New member
just checking. do you mean ...

$Z = \left(\dfrac{1}{5} j - 1 \right) + \left(\dfrac{1}{2} j + 6 \right) + \left(\dfrac{1}{4} j \right)$

or ...

$Z = \dfrac{1}{5j - 1} + \dfrac{1}{2j + 6} + \dfrac{1}{4j}$
the second equation

#### skeeter

##### Well-known member
MHB Math Helper
just checking. do you mean ...

$Z = \left(\dfrac{1}{5} j - 1 \right) + \left(\dfrac{1}{2} j + 6 \right) + \left(\dfrac{1}{4} j \right)$

or ...

$Z = \dfrac{1}{5j - 1} + \dfrac{1}{2j + 6} + \dfrac{1}{4j}$
the second equation
That's what I thought. In future, use grouping symbols to set off the denominators like so ...

Z = 1/(5j-1) + 1/(2j+6) + 1/(4j)

... or learn to use Latex.

Meanwhile, a common denominator is needed to add the three expressions ...

$Z = \dfrac{(2j+6)(4j)}{(5j - 1)(2j+6)(4j)} + \dfrac{(5j-1)(4j)}{(5j-1)(2j + 6)(4j)} + \dfrac{(5j-1)(2j+6)}{(5j-1)(2j + 6)(4j)}$

$Z = \dfrac{(24j-8)-(4j+20)+(28j-16)}{(5j-1)(2j + 6)(4j)}$

$Z = -\dfrac{48j - 44}{64j+112}$

can you complete the simplification from here?

#### DavidCampen

##### Member
the second equation
Then, instead of writing "Z = (1/5j-1) + (1/2j+6) + (1/4j)" you should have written something like "Z = 1/(5j-1) + 1/(2j+6) + 1/(4j)" .

Also, are the values impedances or admittances? This can't be determined just by looking at the complex numbers though the "Z = " could be construed to imply that they are impedances.

Also, why do you write the imaginary part of the complex number before the real part? Usually it is written the other way around.