Dividing complex numbersss

  • Thread starter thomas49th
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  • #1
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Homework Statement


The complex number z is defined by

[tex]z = \frac{a+2i}{a-i}[/tex]

Given that the real part of z is 1/2, find the value of a


Homework Equations


The Attempt at a Solution


Well first of all i multiplied the numerator and denominator of z by (a+i)
which gave me

[tex]\frac{a^{2} + 3ai + 1}{a^{2}-1}[/tex]

Now the real part is going to be a² + 1

so i set

[tex]\frac{a^{2} + 1}{a^{2} - 1} = \frac{1}{2}[/tex]

however I get a = [tex]\sqrt{-3}[/tex]

Have I gone the right way about solving this question?

Thanks :)
 

Answers and Replies

  • #2
nicksauce
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The method looks good, however I believe the numerator should be a^2 + 3ai -2, not a^2 + 3ai + 1.
 
  • #3
gabbagabbahey
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and the denominator should be a^2+1 not a^2-1
 
  • #4
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Ahh yes i can see i made a istake in the numerator however I dont see how the denominator can be a² - 1?:
(a-i)(a+i)
= a² + ai - ai - i²
= a² -1

Does i x i always = 1 regardless of what the sign before it is?

Thanks :)
 
  • #5
tiny-tim
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Ahh yes i can see i made a istake in the numerator however I dont see how the denominator can be a² - 1?:
(a-i)(a+i)
= a² + ai - ai - i²
= a² -1

Does i x i always = 1 regardless of what the sign before it is?

Thanks :)

No … (a-i)(a+i)
= a² + ai - ai - i²
= a² + 1.

i x i always = -1 regardless of what the sign before it is! :smile:
 
  • #6
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ahh i remeber asking one of my maths teachers this question

i x i is always -1

but, as surds:

[tex] \sqrt{-1} . \sqrt{-1} = \sqrt{-1 \times -1} = \sqrt{1}[/tex]

does it not?

Thanks
 
  • #7
tiny-tim
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ahh i remeber asking one of my maths teachers this question

i x i is always -1

but, as surds:

[tex] \sqrt{-1} . \sqrt{-1} = \sqrt{-1 \times -1} = \sqrt{1}[/tex]

does it not?

Thanks

hee hee! :biggrin:

nooo … it doesn't work that way …

√ (or 1/2) is ambiguous, like arcsin, and it's not a good idea to use anything ambiguous in a general formula! :smile:
 
  • #8
35,287
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ahh i remeber asking one of my maths teachers this question

i x i is always -1

but, as surds:

[tex] \sqrt{-1} . \sqrt{-1} = \sqrt{-1 \times -1} = \sqrt{1}[/tex]

does it not?

Thanks
The rule for square roots is that [tex]\sqrt{ab} = \sqrt{a}\sqrt{b}[/tex] provided that a and be are nonnegative.
 
  • #10
Mentallic
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hehe thomas49th I like the way you think.

While I always looked at [tex]\sqrt{-1}^2=-1[/tex] since I just cancel the square with the root, I never thought about expressing it your way. I too would've been a little confused if I ever did end up doing it that way. :smile:
 
  • #11
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hehe thomas49th I like the way you think.

While I always looked at [tex]\sqrt{-1}^2=-1[/tex] since I just cancel the square with the root, I never thought about expressing it your way. I too would've been a little confused if I ever did end up doing it that way. :smile:

I hope this is the way you used to do things, but don't do them that way now! Your equation [tex]\sqrt{-1}^2=-1[/tex] is incorrect on at least two counts:
  1. [tex]-1^2=-1[/tex], so you're attempting to take the square root of -1. If you meant the square of -1, rather than the negative of 1 squared, put parentheses around -1.
  2. [tex]\sqrt{(-1)^2}=+1, not -1[/tex]
 
  • #12
Mentallic
Homework Helper
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I hope this is the way you used to do things, but don't do them that way now! Your equation [tex]\sqrt{-1}^2=-1[/tex] is incorrect on at least two counts:
  1. [tex]-1^2=-1[/tex], so you're attempting to take the square root of -1. If you meant the square of -1, rather than the negative of 1 squared, put parentheses around -1.
  2. [tex]\sqrt{(-1)^2}=+1, not -1[/tex]

well I meant [tex](\sqrt{-1})^2[/tex]. Of course when I see i2 I instantly think -1, and don't get into the nitty gritty of it :smile:
 

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