How to find z^n of a complex number

[arhzz]

Homework Statement

Find z^6 of a complex number

Relevant Equations

-

Hello! (Not sure if this is pre or post calc,if I am in the wrong forum feel free to move it)

So im given this complex number $z = \frac{6}{1-i}$ and I am susposed to get it in polar form as well as z = a+bi

I did that; z = 3+3i and polar form $z =\sqrt{18} *e^{\pi/4 i}$

Now Im susposed to find $z^4$ I didnt know how to that of the top of my head so I googled a bit and found that this should work.

$$ z^n = r^n(\cos(\phi n) +i\sin(\phi n)) $$ Okay so my n = 4, phi should be 45 degrees. So i typed everything in my calculator and I get -324.

This is wrong,the answer should be $z^4 = -243$ But here is the part that confuses me

There is a very very similar problem to this one; where z is $z = \frac{6}{1+i}$ The exact same is required and I did everything the same. I got z = a+bi and polar form; both are correct according to the solutions.

And I tried the same formula I used here and I get the same result $z^4 = -324$ and this is the correct result.

What am I missing in the first example? Is it just luck that in the second example the solutions match,and I am not doing this right? If this is not the way how should I approach this.Thanks in advance!

 

[Merlin3189]

No expert here, but I get z= 3 +3i and z^4 = -324 same as you.
For the second, z= 6/(1 +i) I get z=3 -3i and z^4 = +324
Maybe the first -243 is a typo?

 

[arhzz]

No expert here, but I get z= 3 +3i and z^4 = -324 same as you.
For the second, z= 6/(1 +i) I get z=3 -3i and z^4 = +324
Maybe the first -243 is a typo?

Hmm I dont know I'm really starting to suspect it might be a typo.But how do you get a positive value for the second one? Could you please post how you are calculating that?

 

[SammyS]

Homework Statement:: Find z^6 of a complex number
Relevant Equations:: -

Hello! (Not sure if this is pre or post calc,if I am in the wrong forum feel free to move it)

So im given this complex number $z = \frac{6}{1-i}$ and I am susposed to get it in polar form as well as z = a+bi

I did that; z = 3+3i and polar form $z =\sqrt{18} *e^{pi/4 i}$

Now I'm supposed to find $z^4$ I didn't know how to that of the top of my head so I googled a bit and found that this should work.

You could write z as $\displaystyle z =3\sqrt{2}\ e^{(\pi/4) i}$ . Right? (Use \pi for $\pi$.)

What is $\displaystyle \left(3\sqrt{2} \ e^{(\pi/4) i}\right)^4$ ?

$$ z^n = r^n(cos(\phi n) +isin{\phi n) $$ Okay so my n = 4, phi should be 45 degrees. So I typed everything in my calculator and I get -324.

This is wrong. The answer should be $z^4 = -243$ But here is the part that confuses me

There is a very very similar problem to this one; where z is $z = \frac{6}{1+i}$ The exact same is required and I did everything the same. I got z = a+bi and polar form; both are correct according to the solutions.

And I tried the same formula I used here and I get the same result $z^4 = -324$ and this is the correct result.

What am I missing in the first example? Is it just luck that in the second example the solutions match, and I am not doing this right? If this is not the way how should I approach this.Thanks in advance!

 

[DaveE]

Yes I agree, as does my calculator which does complex math. The answer is -324 for both.

I guess the real lesson here is to recheck your work and then be confident. Can you bill your instructor for your lost time?

 

[arhzz]

Yes I agree, as does my calculator which does complex math. The answer is -324 for both.

I guess the real lesson here is to recheck your work and then be confident. Can you bill your instructor for your lost time?

To be honest I did not lose so much time on that. I've also checked my work multiple times and did another similar problem and it gives me the correct result (calculating it the same way I did the first 2 problems)

Thanks for your help!

 

[Merlin3189]

Oops, yes. -324 for both. Sorry. (-18i)^2 = (+18i)^2 = -324

 

[SammyS]

Homework Statement:: Find z^6 of a complex number
Relevant Equations:: -

Hello! (Not sure if this is pre or post calc,if I am in the wrong forum feel free to move it)

. . .

But here is the part that confuses me

There is a very very similar problem to this one; where z is $z = \frac{6}{1+i}$ The exact same is required and I did everything the same. I got z = a+bi and polar form; both are correct according to the solutions.

And I tried the same formula I used here and I get the same result $z^4 = -324$ and this is the correct result.

Thanks in advance!

I didn't read the second part of your OP carefully enough.

Also, I assume that the Homework Statement should be: Find $z^4$ of a complex number.

So you first had $\displaystyle z = \dfrac{6}{1-i}$, which can be written as $\displaystyle z=3+3i$. You also found that in this case, $\displaystyle z^4 = -324$ , which is correct, as several people have said.

Then you mentioned another problem, namely, find $z^4$ for $z$ being $\displaystyle \dfrac{6}{1+i}$ . (For this $z$ you should find $\displaystyle z=3-3i$). You seemed puzzled to find that even though $z$ was different, you got the same value,
$-324,$ for $z^4$.

To help answer why they're the same:

What do you get for $\displaystyle (3+3i)^2$ ?

What do you get for $\displaystyle (3-3i)^2$ ?

 

[PeroK]

Alternatively:$$(1-i)^4 = 1 -4i -6 +4i +1 = -4$$$$(1 + i)^4 = 1 +4i -6 -4i +1 =-4$$

 

[arhzz]

Got it now thank you!

 

[pasmith]

What am I missing in the first example? Is it just luck that in the second example the solutions match,and I am not doing this right? If this is not the way how should I approach this.

$1 + i$ and $1 - i = \overline{1 + i}$ are complex conjugates. Since $\overline{z^n} = \bar{z}^n$ for any positive or negative integer $n$, $$\left(\frac{6}{1 + i}\right)^{4} = \left(\frac{6}{\overline{1-i}}\right)^{4} = \overline{\left(\frac{6}{1 - i}\right)^{4}} = \overline{-324} = -324.$$

 

[Mark44]

You could write z as
$\displaystyle z =3\sqrt{2}\ e^{(\pi/4) i}$ . Right? (Use \pi for $\pi$.)

Fixed the issue with $\pi$ in the OP, and edited a subsequent equation that wasn't rendering.

 

[Mark44]

Since $\overline{z^n} = \bar{z}^n$ for any positive or negative integer $n$,

I find it much quicker to write ## than [itex] for inline Tex, and $$ rather than [tex] for standalone Tex.