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# Could somebody help me check this?

+5
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fiora  Apr 21, 2015

### Best Answer

#13
+26961
+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.

.

Alan  Apr 21, 2015
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i stand for sqrt -1 ......and i might not be here aftet a minute....

fiora  Apr 21, 2015
#2
+519
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Men, I had been working on this several days,recently figured this out....

fiora  Apr 21, 2015
#3
+519
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I am going to sleep.See you everybody

fiora  Apr 21, 2015
#4
+93289
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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

Melody  Apr 21, 2015
#5
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FWIW, I got the same answer as fiora.

# ☼☼☼

Badinage  Apr 21, 2015
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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?

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

Melody  Apr 21, 2015
#8
+93289
+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?

Melody  Apr 21, 2015
#9
+93289
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I had a quick look at Wolfram|Alpha  and  I cannot tell whether it agrees with mine or not since mine is not finished :)

Melody  Apr 21, 2015
#10
+93289
+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

Melody  Apr 21, 2015
#11
+889
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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,

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

Bertie  Apr 21, 2015
#12
+520
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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?

Badinage  Apr 21, 2015
#13
+26961
+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
+93289
+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. :)

Melody  Apr 21, 2015
#15
0

Did you write this with your left or right foot?

Guest Apr 21, 2015
#16
+520
+5

Thanks Alan.  .

Badinage  Apr 21, 2015

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