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# (-1)^4/3

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Is it possible to do (-1)^4/3

Apr 25, 2014

### Best Answer

#4
+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

### 6+0 Answers

#1
+115955
+5

Is it possible to do (-1)^4/3

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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
+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
+112477
+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
+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
+112477
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
+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