Complex number question: express -8 - 8sqrt(3) in form of...

[Peter Burnes]

Hi..... Have a teaser that i worked through but i don't think is right. Any ideas?
the question is as follows:

Express -8 - 8SQRT(3) in the form r(cos[theta] + isin[theta])

Working with the Argand diagram, I got an answer in the third quadrant (unit circle) of:

7[pi]/6 for the angle [theta]

It goes on to say:

Hence find (-8 - 8[sqrt]3)³

I used deMoivres Theorem, but I'm not sure if it's right. Any suggestions?

 

[tkhunny]

Sorry, but "8-8SQRT(3)" is a Real Number. Did you mean "8-8SQRT(3)i"? If so, you still don't have it.

Where's your magnitude, r?

It is in Quadrant III, but not at \(\displaystyle \frac{7}{6}\pi\).

DeMoivre is good, but you'll have to start better before you get to that.

Give it another go.

 

[Peter Burnes]

yes it is
8-8SQRT(3)i

The magnitude r i got as being 16

To get teta i calculated the inverse tan of -8sqrt(3)/-8 which gave me the inverse tan of the sqrt(3) which is 30 degrees....wait a sec.... it's 60 not 30 degrees...ok i can see where one of my mistakes is... cheers coach!!
I'll tip away!

 

[Peter Burnes]

ok
i got 4[pi] over 3. in the third quadrant which gives me an angle of 240 for teta.

which is in radians 4[pi] over 3

So would 16(cos 4[pi] over 3 + i sin4[pi] over 3) be correct?

would the second part be ?

(-8 - 8[sqrt]3)^3 = 16(cos (3)4[pi] over 3 + i sin (3) 4[pi] over 3)


thanks for the super quick reply b.t.w!!!!

 

[tkhunny]

Nice try. You missed the magnitude.

 

[Peter Burnes]

the magnitude? Isn't that the square root of the real number to be squared + the imaginary number to be squared? I got 16..i'm fairly sure that's correct.
What are your thoughts?

 

[soroban]

Re: Complex number question: express -8 - 8sqrt(3) in form o

Hello, Peter!

Let's walk through this step-by-step . . .

(a) Express \(\displaystyle \,-8\,-\,8\sqrt{3}i\,\) in the form \(\displaystyle \,r(\cos\theta\,+\,{\bf i}\sin\theta)\)

(b) Hence find: \(\displaystyle \,(-8\,-\,8\sqrt{3}{\bf i})^3\)

(a) The complex number \(\displaystyle \,z\:=\:a\,+\,b{\bf i}\,\) has magnitude: \(\displaystyle \,r\:=\:\sqrt{a^2\,+\,b^2\)

\(\displaystyle \;\;\)and angle \(\displaystyle \theta\), where: \(\displaystyle \,\tan\theta\:=\:\frac{b}{a}\)


We have: \(\displaystyle z\:=\:-8\,-\,8\sqrt{3}{\bf i},\;\;a\,=\,-8,\;\;b\,=\,-8\sqrt{3}\)

Then: \(\displaystyle \,r\;=\;\sqrt{(-8)^2\,+\,(-8\sqrt{3})^2} \;=\;\sqrt{64\,+\,192}\;=\;\sqrt{256}\;=\;16\)

And: \(\displaystyle \,\tan\theta\;=\;\frac{-8\sqrt{3}}{-8}\;=\;\sqrt{3}\;\;\Rightarrow\;\;\theta\;=\;\frac{\pi}{3},\;\frac{4\pi}{3}\)

\(\displaystyle \;\;\)Since \(\displaystyle \theta\) is in quadrant 3: \(\displaystyle \,\theta\,=\,\frac{4\pi}{3}\)


Therefore: \(\displaystyle \,-8\,-\,8\sqrt{3}{\bf i}\;=\;16\left(\cos\frac{4\pi}{3}\,+\,{\bf i}\sin\frac{4\pi}{3}\right)\)


(b) Using DeMoivre's theorem:

\(\displaystyle \;\;\left[16\left(\cos\frac{4\pi}{3}\,+\,{\bf i}\sin\frac{4\pi}{3}\right)\right]^3 \;= \;16^3\left[\cos\left(3\cdot\frac{\4\pi}{3}\right)\,+\,{\bf i}\sin\left(3\cdot\frac{4\pi}{3}\right)\right]\)

\(\displaystyle \;=\;4096\left[\cos(4\pi)\,+\,{\bf i}\sin(4\pi)\right] \;=\;4096\left[1\,+\,0{\bf i}\right] \;= \; 4096\)

 

[Peter Burnes]

Thx so much!!

 

[Peter Burnes]

Calculating four complex numbers

That's a great answer!
I have another quick question that's not really covered in my book at the moment.

Find the FOUR complex numbers such that
z^4=\(\displaystyle \,-8\,-\,8\sqrt{3}i\,\)

I have to give my answer in the form a +bi with a and b fully evaluated.

Do i just apply deMoivres Theorem...I'm not sure how that will give me four complex numbers though...

Thanks for your input!

 

[tkhunny]

See that extra parameter in the numerators? It's usually 'k'. Read its definition very carefully. That will get you all the roots you need.

 

[soroban]

Re: Calculating four complex numbers

Hello, Peter!

Perhaps you don't understand DeMoivre's Theorem ... as it applies to roots.
I'll give you a walk-through . . .

Find the FOUR complex numbers such that: \(\displaystyle \,z^4\:=\:-8\,-\,8\sqrt{3}i\)

I have to give my answer in the form \(\displaystyle a\,+\,bi\) with \(\displaystyle a\) and \(\displaystyle b\) fully evaluated.

Do i just apply DeMoivre's Theorem?
I'm not sure how that will give me four complex numbers though

If the polar form is: \(\displaystyle \,z\:=\:r(\cos\theta\,+\,i\cdot\sin\theta)\)

\(\displaystyle \;\;\)then: \(\displaystyle \,z^{\frac{1}{k}} \;=\;r^{\frac{1}{k}}\left[\cos\left(\frac{\theta\,+\,2\pi n}{k}\right)\,+\,i\cdot\sin\left(\frac{\theta\,+\,2\pi n}{k}\right)\right]\;\) . . . for \(\displaystyle n\,=\,0,\,1,\,2,\,\cdots\,k-1\)


We have: \(\displaystyle \,z^4\;=\;16\left(\cos\frac{4\pi}{3}\,+\,i\cdot\sin\frac{4\pi}{3}\right)\) . . . and we want the 4^th roots.

Then: \(\displaystyle \,z\;=\;16^{\frac{1}{4}}\left[\cos\left(\frac{\frac{4\pi}{3}\,+\,2\pi n}{4}\right)\,+\,i\cdot\sin\left(\frac{\frac{4\pi}{3}\,+\,2\pi n}{4}\right)\right]\)

Simplify: \(\displaystyle \,z\;=\;2\left[\cos\left(\frac{\pi}{3}\,+\,\frac{\pi}{2}n\right)\,+\,i\cdot\sin\left(\frac{\pi}{3}\,+\,\frac{\pi}{2}n\right)\right]\;\) . . . for \(\displaystyle n\,=\,0,\,1,\,2,\,3\)


\(\displaystyle n\,=\,0:\;\;z\:=\:2\left(\cos\frac{\pi}{3} + i\cdot\sin\frac{\pi}{3}\right) \:= \;2\left(\frac{1}{2}\,+\,i\frac{\sqrt{3}}{2}\right) \:=\;1\,+\,i\sqrt{3}\)

\(\displaystyle n\,=\,1:\;\;z\:=\:2\left(\cos\frac{5\pi}{6}\,+\,i\cdot\sin\frac{5\pi}{6}\right)\:=\:2\left(-\frac{\sqrt{3}}{2}\,+\,i\frac{1}{2}\right) \:= \:-\sqrt{3}\,+\,i\)

\(\displaystyle n\,=\,2:\;\;z\:=\:2\left(\cos\frac{4\pi}[3}\,+\,i\cdot\sin\frac{4\pi}{3}\right)\:=\:2\left(-\frac{1}{2}\,-\,i\frac{\sqrt{3}}{2}\right) \:=\: -1\,-\,i\sqrt{3}\)

\(\displaystyle n\,=\,3:\;\;z\:=\:2\left(\cos\frac{11\pi}{6}\,+\,i\cdot\sin\frac{11\pi}{6}\right)\:=\:2\left(\frac{\sqrt{3}}{2}\,-\,i\frac{1}{2}\right) \:=\:\sqrt{3}\,-\,i\)
.