## 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

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

gabbagabbahey
Homework Helper
Gold Member
and the denominator should be a^2+1 not a^2-1

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 :)

tiny-tim
Homework Helper
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! 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
Homework Helper
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
Mentor
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.

ah cool :)

Mentallic
Homework Helper
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
Mentor
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
Homework Helper
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 