Author Topic: Complex Number Help! (Read 5588 times) Tweet Share

dcc · « **Reply #30 on:** August 10, 2008, 11:26:30 am »

Quote from: bigtick on August 10, 2008, 10:03:51 am

There is another value for (-1)^(1/2) besides i. Think complex.

Let $a = (-1)^{\frac{n}{2}} = \exp \left(i \left( \pi + 2 \pi m \right) \right)^{\frac{n}{2}} = \exp \left( i \left(\frac{\pi n}{2} + \pi m n \right) \right),\ m \in \mathbb{Z}$

$a = \exp \left(\frac {i \pi n}{2} \right) \cdot \exp \left(i \pi m n\right) = \exp \left(i \pi k\right) \cdot \exp \left( 2 \pi m k \right)$ (since $n = 2k,\ k \in \mathbb{Z}$ )

$\begin{align*} \implies a &= \left( \cos \pi k + i \sin \pi k \right) \cdot \left( \cos 2 \pi m k + i \sin 2 \pi m k \right)\\ a &= \left(\cos \pi k + 0\right) \cdot \left(1 + 0i \right)\\ a &= \cos \pi k = \boxed{ \cos \frac{\pi n}{2} } \end{align*}$

Now, for the other side:

Let $b = i^n = \exp \left( i \left(\frac{\pi}{2} + 2 \pi j\right) \right)^n,\ j \in \mathbb{Z}$

$b = \exp \left( i \left( \frac{\pi n}{2} + 2 \pi n j \right) \right) = \exp \left( \frac{i \pi n}{2} \right) \cdot \exp \left( 2 \pi i n j \right)$

Since $n = 2t,\ t \in \mathbb{Z}$ :

$\begin{align*} b &= \left( \cos \pi t + i \sin \pi t \right) \cdot \left( \cos 4 \pi j t + i \sin 4 \pi j t \right)\\ b &= \left( \cos \pi t + 0 \right) \cdot \left(1 + 0i \right)\\ b &= \cos \pi t = \boxed{\cos \frac{\pi n}{2}}\\ \end{align*} $

So $\boxed{ \left(-1\right)^{\frac{n}{2}} = i^{n} = \cos \frac{\pi n}{2} }$ for even integers, regardless of which branch you select.

Mao · « **Reply #31 on:** August 10, 2008, 12:05:48 pm »

i see four problems with this current debate:

1) we're stuck on the definition of $a^{1/2}$ , which I believe represents the principle branch, but interpreted differently by others.
the habitual thing I have learnt to do is to apply the $\pm$ when a square is removed: $a^2=b\implies a=\pm (b)^{1/2}$
as opposed to counting two branches in a square root [a principle branch evaluation in my terms]: $a=b^{1/2}\implies a=\{-\sqrt{b},\sqrt{b}\}$
*this is also convention used in mathematical computing solutions

2) failing that, $(-1)^{n/2}=(i)^n,\; n\in \mathbb{R}$ is not completely wrong, it represents a branch of solutions. the statement "only for even natural numbers" is right to the same degree as it is wrong.

3) this discussion is way beyond the scope of the question itself. one might argue that the even integers restriction is applied strictly for mathematical correctness, but I would argue that given the context, it was meant simply so that year 11 students can recognise $i^2=-1$

4) [personal opinion] I do not find the attitude and manner in which the disagreement has been voiced to be appropriate nor acceptable. In an intellectual environment such as this, it will be more constructive to refute a statement with reasoning rather than simply state "you are wrong". It does not come friendly, and I have had complaints from others regarding this. Please heed this advice.

Collin Li · « **Reply #32 on:** August 10, 2008, 04:12:22 pm »

If it is to teach $i^2 = -1$ , then it can only be justified for even numbers.

There is no need to be so defensive about what you are doing. It is just worth pointing out that the definition $i$ should not be thought of as $i = \sqrt{-1}$ explicitly, but rather: $i^2 = -1$ (an implicit definition).

Damo17 · « **Reply #33 on:** August 11, 2008, 09:21:17 pm »

Need help with another question my teacher gave me. Tried working out for ages, just can't get it tonight.

Find values for a and b so that $z=a+bi$ satisfies $\frac{z+i}{z+2} = i$

shinny · « **Reply #34 on:** August 11, 2008, 09:40:13 pm »

$\frac{z+i} {z+2}=i$

$\Rightarrow z+i=iz+2i$

$\Rightarrow z-iz=i$

$\Rightarrow z(1-i)=i$

$\Rightarrow z=\frac{i} {1-i}$

$\Rightarrow z=\frac{i(1+i)} {(1-i)(1+i)}$

$\Rightarrow z=\frac{i+i^2} {2}$

$\Rightarrow z=\frac{i} {2} - \frac{1} {2}$

$\Rightarrow a+ib=\frac{i} {2} - \frac{1} {2}$

$\boxed{answer: a=-\frac{1} {2}\; and\; b=\frac{1} {2}}$

first time using latex so hopefully its formatted alright. someone confirm this answer?

dcc · « **Reply #35 on:** August 11, 2008, 09:42:00 pm »

Quote from: Damo17 on August 11, 2008, 09:21:17 pm

Find values for a and b so that $z=a+bi$ satisfies $\frac{z+i}{z+2} = i$

$a + (b + 1)i = - b + (a + 2)i$

So clearly $a = -b$ and $b + 1 = a + 2$

combining these: $b + 1 = -b + 2 \implies \boxed{b = \frac{1}{2},\ a = - \frac{1}{2}}$

edit: whoops lol too late, but same thing

Damo17 · « **Reply #36 on:** August 11, 2008, 09:48:53 pm »

Thanks, fantastic again!!

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Author Topic: Complex Number Help! (Read 5588 times) Tweet Share

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