fraction with complex parts

**suzidoyle635** · March 2nd 2011, 03:03 AM

Hi Im new to complex numbers and I can't figure out this problem.

(j20)(4 + j0.2)/
4 + j(20 + j0.2)

I'm not sure how how to go about doing this to get an answer of 3.77 + j0.944

Can any one help me?

**emakarov** · March 2nd 2011, 03:33 AM

1. Imaginary unit is denoted by i, not j.
2. Are you sure about parentheses? The value of the expression you typed is -1.2 + 40i.
3. If 4 + j(20 + j0.2) is enclosed in parentheses, the answer is still 3.82 + 0.93i, which is different from yours.

**suzidoyle635** · March 2nd 2011, 03:58 AM

on the tutorial it shows some of the working.

(j20)(4 + j0.2)/
4 + j(20 + j0.2)

= (j20)(4 + j0.2)/
4 + j20.2

= (j20)(4 + j0.2)(4-j20.2)/
(4 + j20.2)(4 -j20.2)

=(j20)(4 + j0.2)(4-j20.2)/
16 + 20.2^2
then not sure how to work the numerator out

this is 4 steps, the spacings hasnt worked unfortunately.

**emakarov** · March 2nd 2011, 04:26 AM

If you want to use the fixed-width font and consecutive spaces, you can use the [code]...[/code] tags. However, it is better to write the expression in a single line using parentheses.

Is your expression $\displaystyle\frac{(20i)(4 + 0.2i)}{4 + i(20 + 0.2i)}$ ? Then it is not equal to $\displaystyle\frac{(20i)(4 + 0.2i)}{4 + 20.2i}$ .

**suzidoyle635** · March 2nd 2011, 04:41 AM

the problem initally started as:

1/ + 1/
j15 2.4 + j0.41

Where can I find info on using the maths tags?

**emakarov** · March 2nd 2011, 04:44 AM

Sorry, I don't understand this notation. Please write expressions in a single line using parentheses.

**suzidoyle635** · March 2nd 2011, 04:47 AM

do you mean

(1)/(j15) + (1)/(2.4 + j0.41)

**Soroban** · March 2nd 2011, 04:56 AM

Hello, suzidoyle635!

I assume the original problem was written in Klingon.
No one on this planet writes like that.

Besides, there are obviously some typos . . .

$\dfrac{j20(4 + j0.2)}{4 + j(20 + j0.2)}$

$\text{Answer: }\:3.77 + j0.944$

$\text{I am }guessing\text{ that the problem is: }\;\dfrac{20i(4+0.2i)}{4 + 20i + 0.2i}$

$\text{This simplifies to: }\;\dfrac{\text{-}4 + 80i}{4+ 20.2i}$

$\displaystyle \text{Rationalize: }\;\frac{\text{-}4+80i}{4+20.2i}\cdot\frac{4=20.2i}{4-20.2i} \;=\;\frac{\text{-}16 + 80.8i + 320i + 1616}{16+ 408.04}$

. . . . . . . . $\displaystyle =\;\frac{1600 + 400.8i}{424.04} \;=\;\frac{1600}{424.04} + \frac{400.8}{424.04}i$

. . . . . . . . $=\;3.773228941 + 0.94519385i$

. . . . . . . . $\approx\;3.77 + 0.945i$

**suzidoyle635** · March 2nd 2011, 05:42 AM

Yeah thank you very much for that, problem solved.

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