+0  
 
0
536
6
avatar

Is it possible to do (-1)^4/3 

 Apr 25, 2014

Best Answer 

 #4
avatar+890 
+5

The notation is sloppy isn't it ?

Is it (-1)^4 divided by 3, (and why should the 4 come first, that only happens because the fraction is written on a single line, why not (-1)^(1/3) raised to the power 4 ?), or is it (-1)^(4/3) ?

Enclosing the 4/3 in brackets easily removes the ambiguity, and as this hasn't been done,  my interpretion would be the same as Melody's.

Speaking of Melody, was it yesterday that she asked about

$$e^{i\pi}=-1$$   ?

In which case,

$$(-1)^{4/3}=(e^{i\pi})^{4/3}=e^{4i\pi/3}$$

$$=\cos(4\pi/3)+i\sin(4\pi/3)$$

$$=-\frac{1}{2}-i\frac{\sqrt{3}}{2}.$$

There will of course be two other complex 'answers' one of which is the real number 1.

 Apr 25, 2014
 #1
avatar+111438 
+5

Is it possible to do (-1)^4/3 

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

No, because we would have to use a negative real number as a log base and that creates problems.

Also, note that the normal exponential curve is "unbroken" and non-periodic.

But (-1)2 = 1  and (-1)3 = -1 and (-1)4 = 1........notice how the graph would "oscillate" back and forth between 1 and -1 as x increases in integer value (like a periodic function) and it would also be "undefined" between integer values (a "broken" graph). (That's why you get an "error" on the calculator when you try to evaluate (-1)^(4/3) - your exponent of 4/3 lies "between" integers.)

 Apr 25, 2014
 #2
avatar
+5

Another way to write that question is  (³√-1^4), -1^4=1, ³√1=1

The reason that the calculator puts out errors is it does not simplify the same way.

In short, (-1)^4/3 = 1 , because you are asking for the cube root of -1^4, which is the same as cube root of 1, which is one.

 Apr 25, 2014
 #3
avatar+110163 
+5

$$(-1)^4/3\\\\
=[(-1)^4]/3\\\\
=[1]/3\\\\
=\frac{1}{3}$$

Technically speaking, I belive that is the correct interpretation of the question.  

 Apr 25, 2014
 #4
avatar+890 
+5
Best Answer

The notation is sloppy isn't it ?

Is it (-1)^4 divided by 3, (and why should the 4 come first, that only happens because the fraction is written on a single line, why not (-1)^(1/3) raised to the power 4 ?), or is it (-1)^(4/3) ?

Enclosing the 4/3 in brackets easily removes the ambiguity, and as this hasn't been done,  my interpretion would be the same as Melody's.

Speaking of Melody, was it yesterday that she asked about

$$e^{i\pi}=-1$$   ?

In which case,

$$(-1)^{4/3}=(e^{i\pi})^{4/3}=e^{4i\pi/3}$$

$$=\cos(4\pi/3)+i\sin(4\pi/3)$$

$$=-\frac{1}{2}-i\frac{\sqrt{3}}{2}.$$

There will of course be two other complex 'answers' one of which is the real number 1.

Bertie Apr 25, 2014
 #5
avatar+110163 
0

That is really cool,

Thanks Bertie. 

 

SO now I am a bit confused.

Does (-1)^(4/3) have lots of answers ?

1 and $$\frac{-1-i\sqrt3}{2}$$  are just 2 of them?

 Apr 25, 2014
 #6
avatar+890 
0

The complex number x+iy has two alternative forms, the polar form r(cos(theta)+i sin(theta)) and the exponential form r.e^(i theta), where r is the positive square root of x^2  + y^2  and theta is arctan(y/x). Both polar and exponential form are often abbreviated to r,angle(theta). (Apologies for the text by the way, I can't be bothered fiddling around with LaTex.)

The angle theta is multivalued and strictly should be written theta + 2k.pi, where k is any integer, positive or negative. In most cases the 2k.pi is omitted, but, if the complex number is being raised to a fractional power, it must be included.

Consider e^{i(theta+2k.pi)} raised to the power one third. Using the usual rule for indices this becomes e^{i(theta/3 + 2k.pi/3)} and, as you can see, the angle takes on different values for different values of k. In fact, ignoring the multivaluedness, there will be just three different angles, theta/3, theta/3+2pi/3 and theta/3+4pi/3. The next one theta/3+6pi/3 = theta/3+2pi is a repeat of theta/3. Other values of k simply repeat one of these three angles.

So, there will be three different cube roots of a number, and this applies to any number, not just complex ones. For example, there are three cube roots of 8, 2 angle 0 = 2, 2 angle 2pi/3 = 2(-1/2 + i 3^(1/2)/2) = -1 + i.3^(1/2) and 2 angle 4pi/3 = -1 - i.3^(1/2).

Similarly there will be two square roots (which everyone is used to for positive numbers at least), four fourth roots, five fifth roots and so on.

 Apr 25, 2014

15 Online Users

avatar
avatar