Complex numbers, Homework Question.

In summary, the given expression can be simplified to (5√63+6)j, where j is an imaginary number and j^2= -1. The use of "j" instead of "i" is common among engineers.
  • #1
POLISH_Soldier
5
0
Express this in terms of j


6j-5j2√-63

I have no idea how to do the ones with square roots, my teacher is lost. Completely and I am stuck on this 1 number for like 2 hrs trying to figure it out.

The answer is sopose to be -28j please help me out

**Note that j2= -1

NOTE ALSO THAT the 2 is a 5jsquare
 
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  • #2
You seem to have written the given expression incorrectly. The question itself seems incredibly simple. Can we assume that for "j", you mean j*j= -1 ? Let me try to continue:

[note, you might really want to see the part below starting with "Where..." ]
6j - 5j2(-63)^(0.5)
6j -5*2*j*j(3*3*7)^(0.5)
6j -5*2*3*(-1)*(7^0.5)
6j + 30*7^(0.5)

Where did you find j2 = -1 ? Is that simply a given condition? "2" is a subscript?
6j - 5j2(-63)^(0.5)
6j + 5*(-1)*3*(7^(0.5))
6j - 15*(7^(0.5))
 
  • #3
Oh, those engineers and their jmaginary numbers!

Any normal person would use "i" but those engineers have to use "j"!
I assume you mean 6j-5j^2√-63. If you won't use LaTex at least use ^2 to indicate squares. Yes, J^2= -1 so obviously [itex]6j- 5j^2\sqrt{-3}= 5j\sqrt{63}+ 6j= (5\sqrt{63}+6)j[/itex]. What more is there to say?
 
  • #4
HallsofIvy said:
Oh, those engineers and their jmaginary numbers!

Any normal person would use "i" but those engineers have to use "j"!
I assume you mean 6j-5j^2√-63. If you won't use LaTex at least use ^2 to indicate squares. Yes, J^2= -1 so obviously [itex]6j- 5j^2\sqrt{-3}= 5j\sqrt{63}+ 6j= (5\sqrt{63}+6)j[/itex]. What more is there to say?
It should be [tex]\sqrt{-63}[/tex] in your first equation. You confused the heck out of me for a few minutes :wink:
 

1. What are complex numbers?

Complex numbers are numbers that have two parts: a real part and an imaginary part. They are written in the form a + bi, where a is the real part and bi is the imaginary part with i being the imaginary unit.

2. How are complex numbers used in real life?

Complex numbers are used in a variety of fields, including physics, engineering, and finance. They are used to solve problems involving AC circuits, signal processing, and fluid dynamics, among others.

3. How do you add or subtract complex numbers?

To add or subtract complex numbers, you simply combine the real parts and the imaginary parts separately. For example, (3 + 2i) + (5 + 4i) = (3 + 5) + (2i + 4i) = 8 + 6i. Similarly, (3 + 2i) - (5 + 4i) = (3 - 5) + (2i - 4i) = -2 - 2i.

4. What is the conjugate of a complex number?

The conjugate of a complex number a + bi is a - bi. In other words, the conjugate of a complex number has the same real part but the imaginary part is the opposite sign.

5. How do you multiply or divide complex numbers?

To multiply complex numbers, you can use the FOIL method (First, Outer, Inner, Last) just like multiplying binomials. For division, you multiply the numerator and denominator by the conjugate of the denominator to eliminate the imaginary part in the denominator.

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