Complex number

**Rine198** · April 5th 2011, 03:11 AM

Find the complex number w, lying in the first quadrant, and having the largest possible real part, which satisfies the equation:

w^6=-6-6*I*sqrt(3)

write your answer for w in polar form

Can u please help me with this question? the answer is 12^(1/6)*exp(2/9*I*Pi)

Thanks!

**Plato** · April 5th 2011, 03:29 AM

Originally Posted by Rine198

Find the complex number w, lying in the first quadrant, and having the largest possible real part, which satisfies the equation: w^6=-6-6*I*sqrt(3); the answer is 12^(1/6)*exp(2/9*I*Pi)

If $W=-6-6\sqrt{3}\mathfi{i}$ then $|W|=12$ and $\text{Arg}(W)=-\frac{2\pi}{3}$ .

**Rine198** · April 5th 2011, 03:32 AM

yeah, i got that part, but i still can't seem to get the answer right, so to find the root of w, i divided the argument by 6, but i manage to get -1*Pi/9, and that is not what the answer states

**Plato** · April 5th 2011, 03:36 AM

Originally Posted by Rine198

yeah, i got that part, but i still can't seem to get the answer right, so to find the root of w, i divided the argument by 6, but i manage to get -1*Pi/9, and that is not what the answer states

But the question states that the answer is in the first quadrant.
$\dfrac{2\pi}{6}=\dfrac{3\pi}{9}$ so add.

**Rine198** · April 5th 2011, 03:39 AM

i don't get how u got 2*Pi/6 as well as 3*Pi/9

**Plato** · April 5th 2011, 03:44 AM

Originally Posted by Rine198

i don't get how u got 2*Pi/6 as well as 3*Pi/9

Those six sixth roots are located on a circle centered at 0.
They are equally spaced, separated by $\dfrac{2\pi}{6}.$

**tonio** · April 5th 2011, 03:44 AM

Originally Posted by Rine198

Find the complex number w, lying in the first quadrant, and having the largest possible real part, which satisfies the equation:

w^6=-6-6*I*sqrt(3)

write your answer for w in polar form

Can u please help me with this question? the answer is 12^(1/6)*exp(2/9*I*Pi)

Thanks!

Did you mean to write $w^6=6-6\sqrt{3}i$ above? If so you want the 6-th root

of $6-6\sqrt{3}i=12\,cis(-\pi/3+2k\pi)=12e^{-\pi i/3+2k\pi i}=12e^{5\pi i/3+2k\pi i}$ with the

largest real part , so using de Moivre's formula we get:

$\displaystyle{w_k=\left(12e^{5\pi i/3+2k\pi i}\right)^{1/6}=12^{1/6}e^{\frac{5\pi i}{18}+\frac{2k\pi i}{6}}=12^{1/6}e^{\frac{\pi i}{18}(5+6k)}\,,\,\,k=0,1,2,...,5$ .

Now just write down each root above and check which one has the largest real part. For example,

$Re(w_0)=Re\left(12^{1/6}e^{5\pi i/18}\right)=12^{1/6}\cos\left(5\pi/18)\cong 0.9726$ ,

$Re(w_1)=12^{1/6}\cos\left(11\pi/18)\cong -0.5175$ , etc.

Hint: Of course, you should check only those roots with angle in $\left(-\pi/2\,,\,\pi/2\right) radians$ (why?)

Tonio

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