A Question Involving Complex Numbers

[The Student]

Question: Let z, be an arbitrary complex number. Show z =

if and only if z ∈ R. Here's how I am attempting to solve this. Let

be the complex conjugate of z = a + bi. So if z must be real, then z = a + b, and

= a - bi. a + b = a - bi; b = bi; b^2 = - b^2. I must be misunderstanding what the question is looking for; does anyone have any ideas?

 

[The Student]

No. If z = a + bi is real then b=0.

do this

$\displaystyle z=\overline{z}$

$\displaystyle a+bi=a-bi$

$\displaystyle b=-b$

$\displaystyle b=0$

that's the if direction now you do the only if direction

Here goes: If b does not equal 0 for z = a + bi, then z ∋ ℂ and not an ∋ of ℝ. a + bi = a - bi; b = -b. This is a contradiction. I am not very confident about this.

 

[pka]

Here goes: If b does not equal 0 for z = a + bi, then z ∋ ℂ and not an ∋ of ℝ. a + bi = a - bi; b = -b. This is a contradiction. I am not very confident about this.

Try this. If $\displaystyle b\ne 0$ then $\displaystyle b\ne -b$.

So $\displaystyle a+bi\ne a-bi$ or $\displaystyle z\ne \overline{~z~}$.

 

[The Student]

Try this. If $\displaystyle b\ne 0$ then $\displaystyle b\ne -b$.

So $\displaystyle a+bi\ne a-bi$ or $\displaystyle z\ne \overline{~z~}$.

Your answer is what I was going for. Is my work correct in a roundabout way?

 

[The Student]

Here goes: If b does not equal 0 for z = a + bi, then z ∋ ℂ and not an ∋ of ℝ. a + bi = a - bi; b = -b. This is a contradiction. I am not very confident about this.

Does anyone know if I am on the right track with this answer?

 

[Deleted member 4993]

Question: Let z, be an arbitrary complex number. Show z =

if and only if z ∈ R. Here's how I am attempting to solve this. Let

be the complex conjugate of z = a + bi. So if z must be real, then z = a + b, and

= a - bi. a + b = a - bi; b = bi; b^2 = - b^2. I must be misunderstanding what the question is looking for; does anyone have any ideas?

If and only if (Iff) statements generally need proof in both directions.

 

[The Student]

you're making it much harder than it is.

$\displaystyle b=0$

$\displaystyle z=a+bi=a$

$\displaystyle \bar{z}=a-bi=a$

$\displaystyle a=z=\bar{z}=a$

When you asked for the "only if" part, I tried to show it by showing what happens when b does not equal zero. It was in post #3.

 

[JeffM]

When you asked for the "only if" part, I tried to show it by showing what happens when b does not equal zero. It was in post #3.

I must admit I find your proof fuzzy, which may be my failure, but in any case a proof by contradiction is superfluous.

$\displaystyle z \in \mathbb C \implies z = a + bi\ and\ \bar z= a - bi,\ where\ a,\ b \in \mathbb R.$

If direction

$\displaystyle a + bi = z \in \mathbb R \implies b = 0 \implies a + bi = a - bi \implies z = \bar z.$

Only if direction

$\displaystyle z = \bar z \implies a + bi = a - bi \implies bi = - bi \implies b = 0 \implies z = a + bi = a \in \mathbb R.$

 

[The Student]

I must admit I find your proof fuzzy, which may be my failure, but in any case a proof by contradiction is superfluous.

$\displaystyle z \in \mathbb C \implies z = a + bi\ and\ \bar z= a - bi,\ where\ a,\ b \in \mathbb R.$

If direction

$\displaystyle a + bi = z \in \mathbb R \implies b = 0 \implies a + bi = a - bi \implies z = \bar z.$

Only if direction

$\displaystyle z = \bar z \implies a + bi = a - bi \implies bi = - bi \implies b = 0 \implies z = a + bi = a \in \mathbb R.$

Okay, I think that I was overthinking what the "only if" meant. Thanks Jeff!

 

[The Student]

Thanks everyone, I think I get it now :idea: