Solve 3-7i/2+3i: Imaginary Numbers

[Echo 6 Sierra]

Could someone PULEEZ explain how to work the following equation:

3-7i/2+3i

For the life of me I cannot sqeeze this into my brain!

Thank you in advance.

 

[BoulderHead]

Have you tried multiplying both the Numerator and Denominator by the complex conjugant of the Denominator?
…Don’t forget to use the FOIL method.

 

[enigma]

Huh? That doesn't look like a multiplication problem to me, Boulder.

Echo, when adding and subtracting complex terms (imaginary and real), you just keep the like terms together.

In other words, you add the imaginary parts, and then you add the real parts.

For example:

3+8+5i+7i/4=

(3+8)+(5+7/4)i=

11+(27/4)i

 

[BoulderHead]

We had conversed through PM and Echo 6 Sierra didn't make me aware of this. We treated the problem as;

(3-7i)/(2+3i)

Perhaps Echo 6 Sierra can clarify, as I assumed the parenthesis had been mistakenly omitted. If not, my mistake then.

 

[enigma]

Ah hah.

Gotcha. Parenthesis. Yes, that would be division then, wouldn't it?

 

[Echo 6 Sierra]

Yes, I apologize. I mistakenly omitted them. Also, another apology is in order. I mistakenly posted here instead of the Homework section. Thank you both for your input.

 

[dav2008]

Hm...Did you solve it yet?


But yea, conjugate of the denomitator is how u do it...

OR

You can change it to polar ( i think that's what its called) And then u have
r*cis([0])
r₂*cis([0]₂)
and u just divide the radii and subtract the angles...Not sure tho..i know for multiplication u multiply the radii and add the angles so i would assume u do the opposite for division

 

[Echo 6 Sierra]

Problem solved. Thank you to everyone for your input.