Finding the Value of a in a Complex Number Equation

[thomas49th]

Homework Statement

The complex number z is defined by

z = \frac{a+2i}{a-i}

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

\frac{a^{2} + 3ai + 1}{a^{2}-1}

Now the real part is going to be a² + 1

so i set

\frac{a^{2} + 1}{a^{2} - 1} = \frac{1}{2}

however I get a = \sqrt{-3}

Have I gone the right way about solving this question?

Thanks :)

 

[nicksauce]

The method looks good, however I believe the numerator should be a^2 + 3ai -2, not a^2 + 3ai + 1.

 

[gabbagabbahey]

and the denominator should be a^2+1 not a^2-1

 

[thomas49th]

Ahh yes i can see i made a istake in the numerator however I don't 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 :)

 

[tiny-tim]

Ahh yes i can see i made a istake in the numerator however I don't 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!

 

[thomas49th]

ahh i remeber asking one of my maths teachers this question

i x i is always -1

but, as surds:

\sqrt{-1} . \sqrt{-1} = \sqrt{-1 \times -1} = \sqrt{1}

does it not?

Thanks

 

[tiny-tim]

ahh i remeber asking one of my maths teachers this question

i x i is always -1

but, as surds:

\sqrt{-1} . \sqrt{-1} = \sqrt{-1 \times -1} = \sqrt{1}

does it not?

Thanks

hee hee!

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!

 

[Mark44]

ahh i remeber asking one of my maths teachers this question

i x i is always -1

but, as surds:

\sqrt{-1} . \sqrt{-1} = \sqrt{-1 \times -1} = \sqrt{1}

does it not?

Thanks

The rule for square roots is that \sqrt{ab} = \sqrt{a}\sqrt{b} provided that a and be are nonnegative.

 

[thomas49th]

ah cool :)

 

[Mentallic]

hehe thomas49th I like the way you think.

While I always looked at \sqrt{-1}^2=-1 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.

 

[Mark44]

hehe thomas49th I like the way you think.

While I always looked at \sqrt{-1}^2=-1 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.

I hope this is the way you used to do things, but don't do them that way now! Your equation \sqrt{-1}^2=-1 is incorrect on at least two counts:

1. -1^2=-1, 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. \sqrt{(-1)^2}=+1, not -1

 

[Mentallic]

I hope this is the way you used to do things, but don't do them that way now! Your equation \sqrt{-1}^2=-1 is incorrect on at least two counts:

1. -1^2=-1, 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. \sqrt{(-1)^2}=+1, not -1

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