+0  
 
+5
1019
16
avatar+577 

 Apr 21, 2015

Best Answer 

 #13
avatar+30086 
+10

Look at the step where fiora sets -1 + √3i to 3√(-1 + √3i)3

 

In the next step (-1 + √3i)3 is expanded, correctly, to get 8, but then the wrong cube root of 8 is taken.  Fiora says the cube root of 8 is 2.  If this were the right thing to do here then it would mean that  -1 + √3i equals 2, which would be a little unusual to say the least!

 

The fact is that, in the domain of complex numbers, there are three cube roots of 8, namely 2, -1 - √3i and -1 + √3i.  The only valid root here is the last one.  It isn't valid to jump from -1 + √3i to (-1 + √3i)3 and then jump back to a different root from the one you came from!

 

It's like saying -2 squared is 4, the square root of 4 is 2, so we can replace -2 with 2.

 

.

 Apr 21, 2015
 #1
avatar+577 
0

i stand for sqrt -1 ......and i might not be here aftet a minute....

 Apr 21, 2015
 #2
avatar+577 
0

Men, I had been working on this several days,recently figured this out.... 

 Apr 21, 2015
 #3
avatar+577 
+5

I am going to sleep.See you everybody

 Apr 21, 2015
 #4
avatar+109520 
+5

Hi Fiora,  welcome to the forum.  

Did I talk to you before and suggest that you join and that you take that icon?

It looks great :))

-------------------------------------------------

 

   Edited: this was wrong 

 Apr 21, 2015
 #5
avatar+520 
+10

FWIW, I got the same answer as fiora.

 Apr 21, 2015
 #6
avatar+109520 
+10

Sorry Badinage and fiora,  I think you are right - I was suffering from brain freeze

I have not looked from there on Badinage, is the rest of Fiora's work correct?

 

$$\\(-1+\sqrt3i)^3\\\\
=(\sqrt3i-1)^3\\\\
=[(\sqrt3i)^3]+[3(\sqrt3i)^2(-1)]+[3(\sqrt3i)(-1)^2]+[-1^3]\\\\
=[3\sqrt3*-i]+[-3(3*-1)]+[3(\sqrt3i)]+[-1]\\\\
=[-3\sqrt3i]+[9]+[3\sqrt3i]+[-1]\\\\
=+[9]+[-1]\\\\
=+8$$

.
 Apr 21, 2015
 #8
avatar+109520 
+10

ok then, what about 

$$\\(1+\sqrt3i)^3\\\\
=(\sqrt3i+1)^3\\\\
=[(\sqrt3i)^3]+[3*(\sqrt3i)^2]+[3*(\sqrt3i)]+1\\\\
=(3\sqrt3*-i)+(3*3*-1)+(3*\sqrt3i)+1\\\\
=(-3\sqrt3i)+(-9)+(3\sqrt3i)+1\\\\
=-8$$

 

SO

$$\\\sqrt[3]{ \frac{14*\sqrt[3]{(1+3i)^3} }{2}}\\\\
=\sqrt[3]{ \frac{14*\sqrt[3]{-8} }{2}}\\\\
=\sqrt[3]{ \frac{14*\sqrt[3]{8}*\sqrt[3]{-1} }{2}}\\\\
=\sqrt[3]{ \frac{14*2*\sqrt[3]{-1} }{2}}\\\\
=\sqrt[3]{ 14\sqrt[3]{-1} } \\\\$$

 

I have yet to consider   $$\sqrt[3]{-1}$$

 

Does that look right or wrong to you Badinage?

 Apr 21, 2015
 #9
avatar+109520 
+5

I had a quick look at Wolfram|Alpha  and  I cannot tell whether it agrees with mine or not since mine is not finished :)

 Apr 21, 2015
 #10
avatar+109520 
+10

Now I have to consider

 

  $$\\\sqrt[3]{-1}\\
$The first obvious answer is $ -1+0i $ \;\;since (-1)^3=-1\\
$Now, there will be 3 roots$\\
$each will be $2\pi/3 $ radians apart. $\\
$So the other 2 will be $\\
=cos(\pi/3)+isin(\pi/3)\qquad and \qquad cos(-\pi/3)+isin(-\pi/3)\\
=\qquad \frac{1}{2}+\frac{\sqrt3i}{2} \qquad \qquad \qquad and\qquad \qquad \frac{1}{2}-\frac{\sqrt3i}{2}\\
=\qquad \frac{1+\sqrt3i}{2} \qquad \qquad \qquad and\qquad\qquad\frac{1-\sqrt3i}{2}$$

 

NOW

$$\\\sqrt[3]{-1}=-1,\qquad \frac{1+\sqrt3i}{2}, \qquad\frac{1-\sqrt3i}{2}\\
so\\
\sqrt[3]{14\sqrt[3]{-1}}\\\\
=\sqrt[3]{14(-1)},\qquad \sqrt[3]{14*\frac{1+\sqrt3i}{2}}, \qquad \sqrt[3]{14*\frac{1-\sqrt3i}{2}}\\\\
=\sqrt[3]{-14},\qquad \sqrt[3]{7(1+\sqrt3i)}, \qquad \sqrt[3]{7(1-\sqrt3i)}\\\\
$Now WolframAlpha only gave 3 answers but here I am going to get 3 lots of 3 answers, perhaps each lot is the same as the other 3 lots?$\\\\
\sqrt[3]{-14}\\\\
=\sqrt[3]{14}\sqrt[3]{-1}\\\\
=-\sqrt[3]{14},\qquad \sqrt[3]{14}*\frac{1+\sqrt3i}{2}, \qquad\sqrt[3]{14}*\frac{1-\sqrt3i}{2}\\\\$$

 

$$\\$This approximates to$\\\\
=-2.410,\qquad 2.410*\frac{1+\sqrt3i}{2}, \qquad2.410*\frac{1-\sqrt3i}{2}\\\\
=-2.410,\qquad 2.410*(0.5+\frac{\sqrt3i}{2}), \qquad2.410*(0.5-\frac{\sqrt3i}{2})\\\\
=-2.410,\qquad 1.205+2.087i, \qquad 1.205-2.087i\\\\$$

 

Badinage is right.   None of these answers agree with Wolfram|alpha     

 Apr 21, 2015
 #11
avatar+890 
+10

The best method for dealing with powers and roots of complex numbers, is to switch them to polar form and then to use De Moivre's theorem.

So here,

$$\displaystyle \frac{-7+21\sqrt{3}\imath}{2}=-\frac{7}{2}(1-3\sqrt{3}\imath)=-\frac{7}{2}\sqrt{28}\angle\tan^{-1}(-3\sqrt{3})$$

(ask for details if you need them)

$$\displaystyle = -7\sqrt{7}\angle 280.89339\deg.$$

Raise that to the power one third and you have,

$$(-7\sqrt{7}\angle(-280.89339+360k))^{1/3}$$

$$\displaystyle = -\sqrt{7}\angle(93.63113+120k), \quad k=0,1,2.$$

That gets you

$$\displaystyle k=0:\quad -\sqrt{7}\angle 93.63113 = -\sqrt{7}(\cos93.63113+\imath\sin93.63113)$$

$$\displaystyle =0.16752-2.64044\imath.$$

and the other two values of k,

$$\displaystyle 2.20291+1.46533\imath \;\text{ and } -2.37047+1.17511\imath \quad\text{respectively}.$$

.
 Apr 21, 2015
 #12
avatar+520 
+5

I'd like someone to point out the flaw in fiora's working. I agree there exist more than one answer, but where is the error in fiora's calculation if the result obtained there is not one of the valid answers?

 Apr 21, 2015
 #13
avatar+30086 
+10
Best Answer

Look at the step where fiora sets -1 + √3i to 3√(-1 + √3i)3

 

In the next step (-1 + √3i)3 is expanded, correctly, to get 8, but then the wrong cube root of 8 is taken.  Fiora says the cube root of 8 is 2.  If this were the right thing to do here then it would mean that  -1 + √3i equals 2, which would be a little unusual to say the least!

 

The fact is that, in the domain of complex numbers, there are three cube roots of 8, namely 2, -1 - √3i and -1 + √3i.  The only valid root here is the last one.  It isn't valid to jump from -1 + √3i to (-1 + √3i)3 and then jump back to a different root from the one you came from!

 

It's like saying -2 squared is 4, the square root of 4 is 2, so we can replace -2 with 2.

 

.

Alan Apr 21, 2015
 #14
avatar+109520 
+5

Thanks Bertie, I shall have to take time to look at your answer  :)

Thanks for that great explanation Alan, I had vagely thought there could be problems but your explanation clarified it for me.   :)

Thank also Badinage.  Your extra inputs really helped my thoughts. :)

 Apr 21, 2015
 #15
avatar
0

Did you write this with your left or right foot?

 Apr 21, 2015
 #16
avatar+520 
+5

Thanks Alan.  .

 Apr 21, 2015

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