# [Solved] Simplifying imaginary expression...

#### ballsystemlord

Joined Nov 19, 2018
167
Hello,
I was reading my math text book and came upon a portion about complex numbers. "No problem," I thought, "I did this in algebra 2." Well, algebra 2 was some time ago, and I never used it. Now I'm stuck.
Now they're using "j" for what was called "i" in my algebra book. It's equal to: $\sqrt{-1}$ The problem is that, in the first image, they say that $\frac{1}{j} == -j$ IIRC, $\frac{1}{j} == j^{-1}$
In the cases of the second and third images, I have no idea what they're doing in problem d. They appear to ignore the fact that they're square rooting the whole expression. They seem to assume that the sqrt symbol applies just to the top portion of the expression. I thought, "Maybe the mathematical result is equal between the two." But that's not the case at all according to my TI calculator. Is this just a typo, or am I missing something?

Thanks!

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#### WBahn

Joined Mar 31, 2012
30,294
The symbol 'i' is typically used in math-centric fields, while 'j' is used in EE-centric fields (and many other engineering disciplines) to avoid confusion with the extremely common use of 'i' for electrical current.

$$j \; = \; \sqrt{-1} \\ \frac{1}{j} \; = \; \frac{1}{j}\frac{j}{j} \\ \frac{1}{j} \; = \; \frac{1\cdot j}{j \cdot j} \\ \frac{1}{j} \; = \; \frac{j}{\sqrt{-1} \cdot \sqrt{-1}} \\ \frac{1}{j} \; = \; \frac{j}{\left( \sqrt{-1} \right)^2} \\ \frac{1}{j} \; = \; \frac{j}{-1} \\ \frac{1}{j} \; = \; \frac{j}{-1} \frac{-1}{-1} \\ \frac{1}{j} \; = \; \frac{-1 \cdot j}{-1 \cdot -1} \\ \frac{1}{j} \; = \; \frac{-1 \cdot j}{1} \\ \frac{1}{j} \; = \; -1 \cdot j \\ \frac{1}{j} \; = \; -j \\$$

As for (d), that's a typo -- possibly the person that prepared the solution didn't look carefully enough (may have been a grad student working from handwritten drafts from the author), or the problem might have changed and the solution didn't get updated. The answer should be sqrt(j),

#### WBahn

Joined Mar 31, 2012
30,294
There are several ways to understand your first query.

First, convince yourself of the following:

$$j^2 \; = \; -1 \\ \left( -1 \right)^2 \; = \; 1 \\ \left( j^2 \right)^2 \; = \; 1 \\ \therefore j^4 \; = \; 1$$

Now consider the following:

$$\frac{1}{j} \; = \; j^{-1} \\ \frac{1}{j} \; = \; j^{-1} \cdot 1 \\ \frac{1}{j} \; = \; j^{-1} \cdot j^{4} \\ \frac{1}{j} \; = \; j^{3} \\ \frac{1}{j} \; = \; j^{2} \cdot j \\ \frac{1}{j} \; = \; -1 \cdot j \\ \frac{1}{j} \; = \; -j \\$$

#### ballsystemlord

Joined Nov 19, 2018
167
Thanks, I followed your explanations up to this point.

$$\frac{1}{j} \; = \; j^{3} \\ \frac{1}{j} \; = \; j^{2} \cdot j \\ \frac{1}{j} \; = \; -1 \cdot j \\ \frac{1}{j} \; = \; -j \\$$
How does j end up on the bottom of the expression? What became of the -1?

Thanks!

#### BobTPH

Joined Jun 5, 2013
9,264
Which step are you questioning?

#### WBahn

Joined Mar 31, 2012
30,294
Thanks, I followed your explanations up to this point.

$$\frac{1}{j} \; = \; j^{3} \\ \frac{1}{j} \; = \; j^{2} \cdot j \\ \frac{1}{j} \; = \; -1 \cdot j \\ \frac{1}{j} \; = \; -j \\$$
How does j end up on the bottom of the expression? What became of the -1?

Thanks!
What do you mean how does j end up at the bottom of the expression? The first line starts with what you asserted in your first post, namely that

$$\frac{1}{j} \; = \; j^{-1}$$

I assumed that you were comfortable with this. If that's not the case, then we need to step back and start from a different point. Otherwise, j started out, on the left-hand side, in the denominator and it never moved. All manipulations take place on the right-hand side of the equation.

As for the -1, it is still there.

Do either of the following give you any issues? If so, which one(s)?

$$-1 \cdot 5 \; = \; -5 \\ -1 \cdot x \; = \; -x$$

If you are okay with these, then just replace the 5 or the x with j.

#### MrAl

Joined Jun 17, 2014
11,693
Hello,
I was reading my math text book and came upon a portion about complex numbers. "No problem," I thought, "I did this in algebra 2." Well, algebra 2 was some time ago, and I never used it. Now I'm stuck.
Now they're using "j" for what was called "i" in my algebra book. It's equal to: $\sqrt{-1}$ The problem is that, in the first image, they say that $\frac{1}{j} == -j$ IIRC, $\frac{1}{j} == j^{-1}$
In the cases of the second and third images, I have no idea what they're doing in problem d. They appear to ignore the fact that they're square rooting the whole expression. They seem to assume that the sqrt symbol applies just to the top portion of the expression. I thought, "Maybe the mathematical result is equal between the two." But that's not the case at all according to my TI calculator. Is this just a typo, or am I missing something?

Thanks!
Hi,

Whenever you have a complex number in the denominator (j alone is still complex if you think of the real part as simply being zero) the general procedure for simplifying the expression is to turn the denominator into a real number only while not worrying about if the numerator is complex or not. That is so you can get it into the form:
(a+b*j)/D
where D is the real denoninator.

The general rule is you just multiply BOTH the top and bottom by the complex conjugate of the denominator. The complex conjugate is just the same complex number with the sign of just the imaginary part flipped (minus to plus or plus to minus).
For example the complex number:
a+b*j
turns into:
a-b*j
and the complex number
a-b*j
turns into:
a+b*j
So the complex conjugate is very easy to calculate.

Now given a real numerator and complex denominator:
c/(a+b*j)

we can separate this into the numerator N and the denominator D:
N=c
D=a+b*j
The complex conjugate of D is:
C=a-b*j
then we multiply the denominator D by C:
D*C= a^2-b^2*j^2
and since j^2=-1 that becomes the new denominator and is entirely real:
Dn=a^2+b^2
Note there is no imaginary part there.
Next we multiply the numerator N by the complex conjugate of the denominator also:
N*(a-b*j)
and we get:
c*a-c*b*j
So now we have both the top and bottom multiplied by the complex conjugate so we can form the new number:
(c*a-c*b*j)/(a^2+b^2)
and this allows us to separate the real and imaginary parts into:
real: c*a/(a^2+b^2)
imag:-c*b/(a^2+b^2)
Note both the real and imag parts are both real but when we write it out as one number we must include the 'j':
real+j*imag.

For your example the denominator is just j, so the complex conjugate is just -j (no real part).
Now we multiply the top and bottom by this -j and end up with:
(1*-j)/(j*-j)
and we get:
-j/(-(j^2))
and since j^2 is -1 we get:
-j/(-(-1))
and since the denominator is now just 1 we end up with:
-j/1
which of course is equal to -j.

This is the most general procedure I think and works with any complex number no matter what the numerator and/or denominator happen to be. The real goal is to get the denominator to be entirely real so we can later separate the real and imaginary parts if needed.

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