Complex numbers expression

[mathwannabe]

Hello everybody

I haven't been posting for a pretty long time... I was living my "math books" monastery

So, I felt like I was ready to take some peeks at some of the previous tests for getting into math faculty. I am actually doing pretty good. But there is one problem I did and I am not really sure if my answer is the correct one:

1) [(1 + i)/sqrt(2)]^1996 + [(1 - i)/sqrt(2)]^1996 =

([(1 + i)^2]^998) / ([(sqrt(2))^2]^998) + ([(1 - i)^2]^998) / ([(sqrt(2))^2]^998) =

[(2i)^998] / [2^998] + [(-2i)^998] / [2^998] =

(2^998 * i^998) / (2^998) + ((-2)^998 * i^998) / (2^998) =

i^998 - i^998 = 0

Can someone please just confirm this if it is correct or tell me where I'm wrong if it is not correct. Thanks.

Sorry for the mess... for some reason, latex just doesn't work for me anymore

 

[masters]

Hello everybody

I haven't been posting for a pretty long time... I was living my "math books" monastery

So, I felt like I was ready to take some peeks at some of the previous tests for getting into math faculty. I am actually doing pretty good. But there is one problem I did and I am not really sure if my answer is the correct one:

1) [(1 + i)/sqrt(2)]^1996 + [(1 - i)/sqrt(2)]^1996 =

([(1 + i)^2]^998) / ([(sqrt(2))^2]^998) + ([(1 - i)^2]^998) / ([(sqrt(2))^2]^998) =

[(2i)^998] / [2^998] + [(-2i)^998] / [2^998] =

(2^998 * i^998) / (2^998) + ((-2)^998 * i^998) / (2^998) =

i^998 - i^998 = 0 <====== Should be \(\displaystyle i^{998} + i^{998} = -2 \)

Can someone please just confirm this if it is correct or tell me where I'm wrong if it is not correct. Thanks.

Sorry for the mess... for some reason, latex just doesn't work for me anymore

Hi mathwannabe,

If you take a negative number to an even power, the result is positive. So the sign didn't change there at the end.

The final answer is -2.

 

[pka]

1) [(1 + i)/sqrt(2)]^1996 + [(1 - i)/sqrt(2)]^1996 =

([(1 + i)^2]^998) / ([(sqrt(2))^2]^998) + ([(1 - i)^2]^998) / ([(sqrt(2))^2]^998) =

[(2i)^998] / [2^998] + [(-2i)^998] / [2^998] =

(2^998 * i^998) / (2^998) + ((-2)^998 * i^998) / (2^998) =

i^998 - i^998 = 0

\(\displaystyle \frac{1+i}{\sqrt2}=\cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right)\)

Thus \(\displaystyle \left(\frac{1+i}{\sqrt2}\right)^{1996}=\cos\left(499\pi\right)+i\sin\left(499\pi\right)=-1\)

 

[mathwannabe]

Hi mathwannabe,

If you take a negative number to an even power, the result is positive. So the sign didn't change there at the end.

The final answer is -2.

Yes. I knew I did something wrong. My answer would be true if it was -2^998 instead of (-2)^998.

Thank you, I would probably had it figured out eventually, but my answer was in the list of offered answers so those would be some lost points on the actual exam. I covered a huge chunk of math in a short, short time, so I am still feeling a bit shaky about some things.

 

[mathwannabe]

\(\displaystyle \dfrac{-2^{998} * i^{998}}{2^{998}} = \dfrac{(-1)^{998} * 2^{998} * i^{998}}{2^{998}} = i^{998} \ne - i^{998}.\)

Hello Jeff

I didn't quite catch your post ...

Isn't \(\displaystyle -2^{2n} = -1(2^{2n})\) instead of \(\displaystyle (-1)^{2n}(2^{2n})\)

 

[pka]

Isn't \(\displaystyle -2^{2n} = -1(2^{2n})\) instead of \(\displaystyle (-1)^{2n}(2^{2n})\)

That is correct!


But this question is designed to test polar forms. SO

\(\displaystyle \frac{1+i}{\sqrt2}=\cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right)\)

Thus \(\displaystyle \left(\frac{1+i}{\sqrt2}\right)^{1996}=\cos\left(499\pi\right)+i\sin\left(499\pi\right)=-1\)

AND
\(\displaystyle \frac{1-i}{\sqrt2}=\cos\left(\frac{-\pi}{4}\right)+i\sin\left(\frac{-\pi}{4}\right)\)

Thus \(\displaystyle \left(\frac{1-i}{\sqrt2}\right)^{1996}=\cos\left(-499\pi\right)+i\sin\left(-499\pi\right)=-1\)

The answer is -2.