Is an irrational root of a real number imaginary or real?

We can easily comment the result of a root operation just by the information if the degree of the root is odd or even.

But what if the degree of the root (or power) is irrational?

For example;
$-64 ^ \frac{1}{2} \, = \, j8 \,\,\,\,\, (imaginary)$
$-64 ^ \frac{1}{3} \, = \, -4 \,\,\,\,\, (real)$

$+7^{\pi - 3} \, = \, 7^{0.14159265...}$
Is it real, imaginary or complex?

disregardthat
$$a^r=e^{r Log a}$$ is a root for this exponent and is real for all positive a. Hence 7^pi has a real root. For negative a however we have $$\log a = (2\pi k+\pi) i + Log |a|$$, so $$a^r=e^{r(2\pi k+\pi) i+r Log |a|}=|a|^re^{r(2\pi k+\pi) i}$$ which is real only when $$e^{r(2\pi k+\pi) i }$$ is real. There is a real solution whenever $$\sin((2\pi k+\pi)i) = 0$$ has a solution for an integer k, that is whenever r(2k+1) is an integer. Hence for negative a and irrational r you will get that a^r is not real.

Note that you cannot speak of "the" root. If r is an integer there are r roots of which only some may be real. For irrational r you will infinitely many roots, but if a is negative, all of them are non-real.

Last edited:
mathwonk
Homework Helper
2020 Award
Every positive real number has the form e^r for some real number r. Thus for any positive real number a = e^r, and any real number x, we have a^x = (e^r)^x = e^(rx).

Moreover, there is a power series for computing any power of e, i.e.
e^y = 1 + y + y^2/2+ y^3/(2.3) +........ Thus for any real number y, e^y is also real.

And thus for any positive real number a = e^r, and any real number x, then a^x = e^(rx), is also real.

Mark44
Mentor
We can easily comment the result of a root operation just by the information if the degree of the root is odd or even.

But what if the degree of the root (or power) is irrational?

For example;
$-64 ^ \frac{1}{2} \, = \, j8 \,\,\,\,\, (imaginary)$
$-64 ^ \frac{1}{3} \, = \, -4 \,\,\,\,\, (real)$
Neither of the other responders noticed this, so I will comment on it. Your first equation is incorrect.
-641/2 = $-\sqrt{64}$ = -8, a real number.

What you no doubt meant but wrote incorrectly was (-64)1/2, which we math people write as 8i.

disregardthat
Neither of the other responders noticed this, so I will comment on it. Your first equation is incorrect.
-641/2 = $-\sqrt{64}$ = -8, a real number.

What you no doubt meant but wrote incorrectly was (-64)1/2, which we math people write as 8i.

He presumably meant (-64)^(1/2) and wrote j8 where j is the imaginary unit.

Thanks for the correction.
"j" is the imaginary unit for engineer people...

HallsofIvy
Homework Helper
Oh, those engineers and their jmaginary numbers!

$$a^r=e^{r Log a}$$ is a root for this exponent and is real for all positive a. Hence 7^pi has a real root. For negative a however we have $$\log a = (2\pi k+\pi) i + Log |a|$$, so $$a^r=e^{r(2\pi k+\pi) i+r Log |a|}=|a|^re^{r(2\pi k+\pi) i}$$ which is real only when $$e^{r(2\pi k+\pi) i }$$ is real. There is a real solution whenever $$\sin((2\pi k+\pi)i) = 0$$ has a solution for an integer k, that is whenever r(2k+1) is an integer. Hence for negative a and irrational r you will get that a^r is not real.

Note that you cannot speak of "the" root. If r is an integer there are r roots of which only some may be real. For irrational r you will infinitely many roots, but if a is negative, all of them are non-real.

Can I conclude the following equation from your text?

$ln(z)\,=\,ln(re^{j\theta})\,=\,ln(r)\,+\,j(2\pi k\,+\,\theta)$

disregardthat
Can I conclude the following equation from your text?

$ln(z)\,=\,ln(re^{j\theta})\,=\,ln(r)\,+\,j(2\pi k\,+\,\theta)$

These are the values for the multivalued function ln(z). You would need a rule for choosing k i order to get a function out of this, and that is what we call branches for multivalued functions like log. We usually use the principal value http://en.wikipedia.org/wiki/Complex_logarithm#Definition_of_principal_value.

Here is my code, looks like to be working fine for now:

These are working OK:
Code:
long double Math::exp(long double num)
{
long double Ret = 0;
//long double f, p;
for (uint8_t i=0; i<=m_PRECISION_TAYLOR_FACTORIAL; i++)
{
Ret += UnsignedIntegerPower(num, i) / fact(i);
}
return Ret;
}

long double Math::ln(long double num)
{
long double x_k = 1.0, x_k1;
{
x_k1 = x_k + num * exp(-x_k) - 1.0;
if (abs(x_k1 - x_k) < m_PRECISION_EPSILON) break;
x_k = x_k1;
}
return x_k1;
}

long double Math::pow(long double base, long double exponent)
{
bool bNegateResult;
if ((base < 0.0) && (IsInteger(exponent) == false))
{			// complex result
return 0.0;	// we are not dealing with it
}			// use "ComplexNumber" class for it
else if ((base < 0.0) && IsOdd(exponent))
{
bNegateResult = true;
}
else	// if (((base < 0) && IsEven(exponent)) || (base >= 0))
{
bNegateResult = false;
}
double long Ret = exp(exponent * ln(abs(base)));
return bNegateResult ? -Ret : Ret;
}

long double Math::root(long double base, long double degree)
{
return pow(base, static_cast<long double>(1.0) / degree);
}

I haven't test ComplexNumber methods yet:
Code:
ComplexNumber::ComplexNumber(	long double dbRealPartOrRadius/* = 0.0*/,
long double dbImaginaryPartOrAngle/* = 0.0*/,
{
{
}
else
{
m_dbImaginaryPart = dbImaginaryPartOrAngle;
}
}

ComplexNumber ComplexNumber::Power(ComplexNumber cn, long double dbDegree)
{
//theta = Math::GetRreferenceAngle(dbDegree * cn.GetAngle());
Math::GetRreferenceAngle(dbDegree * cn.GetAngle()),
true);
return Ret;
}

std::vector<ComplexNumber> ComplexNumber::FindRoots(ComplexNumber cn, uint64_t nDegree)
{
std::vector<ComplexNumber> Roots;
ComplexNumber FirstRoot = Power(cn, 1.0 / static_cast<long double>(nDegree));
}