Two solutions to complex equation causing a problem

Question

If we look at the equation \begin{align} z = \sqrt{ 8 - 6 i }, \end{align} we will find the solutions \begin{align} z_1 = -3+i \end{align} \begin{align} z_2 = 3 - i \end{align}

How can they be both correct if we can substitute \begin{align} \sqrt{ 8 - 6 i } = -3+i \end{align} \begin{align} \sqrt{ 8 - 6 i } = 3-i \end{align} and if \begin{align} \sqrt{ 8 - 6 i } = \sqrt{ 8 - 6 i } \end{align} then \begin{align} 3-i = -3+i \end{align} and that is of course false. I can't work out where I'm going wrong.

Answer 1

It isn't right to write $\sqrt{8-6i}=3-i$ (the same for the other solution). When you ask, $\sqrt{z}=?$ you mean to say " what thing multiplied by itself gives $z$". It just so happens that this question will not yield an answer that is a single number, but the "thing" is the elements present in the set comprised of two numbers. So what is meant as the solution for this problem is $$z=\sqrt{8-6i} \implies z=\pm(3-i)$$ Some problems even have an infinite number of solutions (even in the real domain), such as trigonometric equations. For example, $\sin{x}=0 $ has an infinite number of solutions on the number line, $x=k\pi$, $k\in \mathbb{Z}$, but it isn't true that they're all the same just because their output through the function "$\sin{x}$" is the same number.
Moreover you can see that no contradiction arises, because $(\pm(3-i))^2=8-6i$ (which doesn't mean $3-i=-3+i$.

Answer 2

This is because taking square roots in the complex plane is not an injective function. In the case of positive real numbers, the square root function is uniquely defined as the positive square root of the number. Here is a quick explanation of branches in the complex plane: https://plus.maths.org/content/maths-minute-choosing-square-roots

Answer 3

Every nonzero complex number $w$ has exactly two distinct square roots (say $z_1$ and $z_2$, so that $z_1^2 = z_2^2 = w$), and they are opposites (i.e., $z_2 = -z_1$). These roots certainly aren't equal; what are equal are their squares. Your mistake is calling them both $\sqrt{w}$. It might be more correct to speak of "a square root" instead of "the square root", since there is more than one square root for each nonzero complex number.

With complex numbers, you have to be careful about what you mean by $\sqrt{w}$; it will be one of $z_1$ or $z_2$, but you have to decide which is which. If you choose to say that $\sqrt{w} = z_1$, then you have immediately also determined that $z_2 = -\sqrt{w}$.

What is really going on is that you are choosing a branch of the logarithm, and using the definition that $\sqrt{w} = e^{\frac12\log w}$. This means that you choose a definition of the argument of $w$, so then $\log w = \ln|w| +i\arg w$ (where $\ln$ is the ordinary real-valued function of a positive real variable).

There are several commonly used ranges for $\arg w$:

• $(-\pi,\pi]$, so $\sqrt{w}$ has argument in $(-\pi/2,\pi/2]$
• $[0,2\pi)$, so $\sqrt{w}$ has argument in $[0,\pi)$

In the first case, $\sqrt{w}$ would lie in the right open half-plane or on the positive $y$-axis, so its opposite $-\sqrt{w}$ would lie in the left open half-plane or on the negative $y$-axis.

In the second case, $\sqrt{w}$ would lie in the upper open half-plane or on the positive $x$-axis, so its opposite $-\sqrt{w}$ would lie in the lower open half-plane or on the negative $x$-axis.

Other choices for defining $\arg w$ would lead to other results for which of the two roots is called $\sqrt{w}$ and which is called $-\sqrt{w}$.

Answer 4

This is precisely why the use of the radical function is restricted to positive real numbers. When looking for the square rootS (notice the plural) of a complex number $a$, you actually look for the solutionS of the following equation : $$ \boxed{z^2 = a}$$

For $a \neq 0$, there are exactly $2$ distinct solutions to the equation. Notice that if you already found a solution $z_1$, then $z_2 := - z_1$ is a solution as well since it satisfies ${z_2}^2 = (-z_1)^2 = {z_1}^2 = a$.

This is why the equation above does not substitute into $z = \sqrt{a}$.

Writing "$z^2 = a \Rightarrow z = \sqrt{a}$" is hence not a valid argument.

For example :

• $a > 0$ a positive real, eg $a = 2$
  The equation $z^2 = 2$ has two solutions : $\sqrt{2}$ and $- \sqrt{2}$.
• $a < 0$ a negative real, eg $a = - 2$
  The equation $z^2 = - 2$ has two solutions : $\sqrt{2} \cdot i $ and $- \sqrt{2} \cdot i$.
• $a \notin \mathbb{R}$ a strictly complex number, eg $a = 8 - 6 \cdot i$
  The equation $z^2 = 8 - 6 \cdot i$ has two solutions : $$\boxed{-3 + i \textbf{ and } 3 - i}$$

All in all, remember that $\sqrt{a}$ only "works" if $a$ is a positive real number, and even so it provides you anly one of the two solutions of $z^2 = a$, namely the positive one.