Simplify this fraction containing imaginary numbers

[serendipityfox]

Homework Statement

1-2i+3i^2 / 1+2i-3i^2 =

a) 3/5 - 1/5i
b) -3/5 + 1/5i
c) -3/5 - 1/5i
d) 3/5 + 1/5i

Relevant Equations

i= i ,i^2= -1

i can get to 3i+1/1-3i but no further. I take it this is the correct way to start

 

[BvU]

Homework Statement: 1-2i+3i^2 / 1+2i-3i^2

That's a complex number, not a homework problem statement (and you omitted the brackets )! What does the composer of the exercise want from you ?

 

[serendipityfox]

there are no brackets in the question... what brackets do you refer to?

 

[BvU]

$1-2i+3i^2 / 1+2i-3i^2 = 1-2i+ {3i^2 \over 1} +2i-3i^2 = 1$ according to the rules of mathematics

 

[PeroK]

Homework Statement: 1-2i+3i^2 / 1+2i-3i^2 =

3i+1/1-3i but no further. I take it this is the correct way to start

That looks wrong in any case.

 

[BvU]

in the question

And we still don't know what the question is

 

[serendipityfox]

what i wrote is exactly what was written on the test paper, there is no more information :(

 

[PeroK]

Homework Statement: 1-2i+3i^2 / 1+2i-3i^2 =

This is considered sloppy notation. It's better (some would say essential) to use brackets:

$(1-2i+3i^2)/(1+2i-3i^2)$

 

[BvU]

Anyway, let's assume (Ass U Me) you want to simplify the expression. Use your own equation for starters -- not the one that says $i=i$ (??) , but

$i^2= -1$

 

[serendipityfox]

@PeroK- thankyou, i understand
@ BvU- this is the first thing i tried, leading to...
(1-2i-3) / (1+2i+3) therefore (-2i-2) / (2i+4) how to proceed?

 

[PeroK]

@PeroK- thankyou, i understand
@ BvU- this is the first thing i tried, leading to...
(1-2i-3) / (1+2i+3) therefore (-2i-2) / (2i+4) how to proceed?

If in doubt, multiply by the complex conjugate!

 

[PeroK]

Homework Statement: 1-2i+3i^2 / 1+2i-3i^2 =

a) 3/5 - 1/5i
b) -3/5 + 1/5i
c) -3/5 - 1/5i
d) 3/5 + 1/5i

Just to go back to the first point. Is $1/5i = \frac{1}{5i}$ or is $1/5i = \frac{i}{5}$

Do you not see the problem? How would anyone know what you really mean?

 

[George Keeling]

what i wrote is exactly what was written on the test paper, there is no more information :(

Can you post a photo of that question?

 

[symbolipoint]

@PeroK- thankyou, i understand
@ BvU- this is the first thing i tried, leading to...
(1-2i-3) / (1+2i+3) therefore (-2i-2) / (2i+4) how to proceed?

THINK! You KNOW this.
If you do not see, then rest for a few minutes or so; and then come back to it. You should then be able to finish in less than two minutes.

 

[symbolipoint]

(1-2i+3i^2) /( 1+2i-3i^2 )

The important 'hint' was given, i^2=-1.

Meaning, (1-2i-3)/(1+2i+3)

and then, (-2-2i)/(4+2i).

NOW, multiply numerator and denominator by the conjugate of the denominator. Simplify from there.

 

[BvU]

(1-2i-3) / (1+2i+3) therefore (-2i-2) / (2i+4) how to proceed?

Your choices are in the form $x+iy$, so you have to get rid of the complex denominator.
To get a real denominator, you multiply by $1\ \$ !.

Spoiler

$$ 1 = {2i-4\over 2i-4} $$ the complex conjugate @symbolipoint was hinting at.
Capisce ?

@moderators: am I giving away too much ?

 

[serendipityfox]

got it, rationalise denominator, i wasn't used to applying it to complaex numbers

 

[SammyS]

got it, rationalise denominator, i wasn't used to applying it to complex numbers

This thread is unnecessarily long, principally due poor mathematical notation as well as problems with the problem statement in general.

When starting a new thread, always include the entire problem statement in the body of the Original Post, no matter what you place in the thread title.

In the case of this thread, the portion of the problem statement found in the thread title could be better. If you quoted this directly from your test paper, then that's not your fault. A better instruction would be something like:

Write the following expression as a complex number in standard form.​

As for the given expression: If it was written on the test as a fraction all on one line, that's unforgivable on the part of the test writer. However, I suspect it may have been given to you as a "stacked" fraction such as:

$\dfrac{1-2i+3i^2 }{ 1+2i-3i^2 }$​

To write this as an equivalent "inline" fraction, you need to enclose. each of the numerator and the denominator in parentheses as follows:

(1 − 2i + 3i^2) / (1 + 2i − 3i^2)

 

[symbolipoint]

Often students are either unaware or not clear-minded about putting in mathematical notation using pure text. Too many things are too new to them. Some students may go through their four years of Mathematics courses in high school and only know how to use the proper notation as done on paper with pencil. Some of us NEVER KNEW about mathematics done in pure text until our first introductory programming course, which might have been two, or four, or five years later. Along the drift about typesetting and text based notation should if needed, become a new forum topic. Avoid the drift here.