# The square root of -1 is not equal to the negative of itself?

Could you help me with this? It's seriously wrong, but I cannot prove that it is wrong. Here is my problem:

Solve $-1^{1.5}$.

I used two ways:

A. Convert $-1^{1.5}$ to $-1^1 * -1^{0.5}$ then solve. The answer is $-i$.

B. Convert $-1^{1.5}$ to $\sqrt {-1^3}$ ($x^{y/z} = \root z \of {x^y}$) then solve. The answer should be $i$.

If $-i = i$, then $-1 = 1$. Would you explain in which step was I wrong?

• $i$ and $-i$ are different numbers. Both of them when squared give $-1$. – Joffan Aug 23 '17 at 1:49
• Using the rules of exponents on negatives bases means you really got to be cautious... – imranfat Aug 23 '17 at 1:51
• $\sqrt{-1^3} = \sqrt{(-1)\times(-1)\times(-1)} = -1\sqrt{-1} = -i$. Is this not the way we do the calculation? And moreover, $x^{y/z} \neq \sqrt[y]{x^z}$. Instead, $x^{y/z} = \sqrt[z]{x^y}$ – Aniruddha Deshmukh Aug 23 '17 at 1:51
• Sorry! I accidentally exchanged those two variables y and z while typing. – Hyunsoo Kim Aug 23 '17 at 1:53
• Note that the square root of $-1$ is either positive or negative $i$. Picking the positive answer to any square root process is an arbitrary choice. With that in mind, both methods get both correct answers. – Kaynex Aug 23 '17 at 1:54

Taking a non-integer power of a complex number may result in more than one answer. In your case, $-1 = e^{i (\pi + 2\pi k)}$, where $k$ is an integer. Consequently, $$(-1)^{1.5} = e^{i 3\pi/2 + 3i \pi k}=\left\{\begin{matrix} -i, & \text{for even k;} \\i, & \text{for odd k.}\end{matrix}\right.$$
In the complex numbers $a^b$ is defined as $e^{b \log a}$ but the complex logarithm is multivalued. For any $z \in \Bbb C, k\in \Bbb Z, e^{z+2k\pi i}=e^z$. In the reals we take the square root sign to return the positive square root, which leads us to put $\pm$ signs in when we take a square root sometimes. In the complex field the positive numbers are not defined, so $-1^{0.5}$ can be either $i$ or $-i$. Similarly, $(-1)^{1.5}$ can be either $i$ or $-i$.