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

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

so the rectangular form will be

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

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

so the rectangular form will be

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

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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 = ${tan}^{-1}(\frac{a}{b})$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?

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"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.

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.

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just checking. do you mean ...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?

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

or ...

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

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the second equationjust 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 ...

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

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 ...

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

That's what I thought. In future, use grouping symbols to set off the denominators like so ...the second equation

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?

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)" .the second equation

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.