# Comparing square roots of negative numbers

If we have for instance $\sqrt{-25}$, that is, a square root of $-25$, I know the answer can be $5i$ (Is $-5i$ also correct? Sorry not professional in mathematics).

My main question here is how to compare square roots of negative numbers.

For example, which is larger, $\sqrt{-11}$ or $\sqrt{-13}$?

Thanks.

• Principal value of $\sqrt{-25}=5i$; you can't compare two complex numbers in general. – Aang Mar 10 '13 at 10:32

## 2 Answers

There is no consistent definition of "larger than" in the complex numbers. One can say that the modulus of $\sqrt{-11}$ is less than that of $\sqrt{-13}$.

• Can you kindly clarify your point: `the modulus of sqrt(-11) is less than that of sqrt(−13)? Thanks – Simplicity Mar 10 '13 at 12:53
• A complex number $a+bi$ with $a$ and $b$ real can be modeled by the point with coordinates $(a,b)$ in the plane. The modulus of $a+bi$ is defined to be the distance, $\sqrt{a^2+b^2}$, from that point to the origin. Exercise: find the modulus of $\sqrt{-11}$; find the modulus of $\sqrt{-13}$; compare them. – Gerry Myerson Mar 10 '13 at 22:05

There is no order relation on $\mathbb C$ (that is compatible with addition and multiplication the way we "need" it). For example, the product of two negative or two positive numbers in an ordered field is always positive. Hence the square of any nonzero number is positive. Hence $i^2=-1$ and $1^2=1$ are both positive and so is their sum $0$ - contradiction.

When talking about the square root function, we must specify which value to return. The answer is not simply that $\sqrt a$ is "any solution of $x^2=a$", as that is not unique (unless $a=0$). Even in the real numbers we define the square root to be the positive solution of $x^2=a$. In the absence of a notion of positiveness in $\mathbb C$, it is customary to define $\sqrt a$ as the solution of $x^2=a$ with positive imaginary part or that is on the nonnegative real axis. This is called the principal value. Please, never write things like $\sqrt{-4}=\pm2i$; it is just as wrong as $\sqrt 4=\pm2$.