# Meaning of the sign of a complex number [closed]

What is the meaning and the practical uses of the sign (signum function) of a complex number $z$ defined as $\frac{z}{|z|}$? Does it also extend to quaternions?

## closed as off-topic by Namaste, Xander Henderson, Did, user21820, user223391 Jul 5 '18 at 13:42

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• Still a valid question. Maybe it doesn't deserve upvotes, but also doesn't deserve downvotes. Downvoting should be kept for bad and invalid questions. – plasmacel Jul 4 '18 at 20:59

If we draw a line from the complex number $z$ to the origin and then we indicate where our line intersects with the unit circle we have identified the point $z/|z|$. Note that when $z$ is real this will simply be $1$ or $−1$.

Generally, we can use this expression $sign$ to mean the direction of the point from the origin. This idea intuitively goes from 1 dimension (+ or -) to $2$ dimensions (any point on the unit circle) to $3$ dimensions (any point on the unit sphere) we can extend it to the quaternians (visualizing the 4th dimension will make my analogy a little harder). People do look at something called unit quarternions and the verser which may be a helpful lead on how this idea extends to 4 dimensions.

You can think of a complex number $z=a+ib$ as a vector in the complex plane, and as such it has a direction and a length. The sign of $z$ gives the normalized direction of $z$ as a unit vector lying on the unit circle. We normalise $z$ by dividing by its modulus: $$\operatorname{sgn}(z)=\frac{z}{|z|}=\frac{a+bi}{\sqrt{a^2+b^2}}$$ This gives a unit complex number. For example take $z=5+3i$. Then $$\operatorname{sgn}(5+3i)=\frac{5+3i}{|5+3i|}=\frac{5}{\sqrt{34}}+\frac{3}{\sqrt{34}}i$$ where $$\left(\frac{5}{\sqrt{34}}\right)^2+\left(\frac{3}{\sqrt{34}}\right)=1$$

For a quaternion $q=a+bi+cj+dk$, the case is similar: $$\operatorname{sgn}(q)=\frac{a+bi+cj+dk}{|a+bi+cj+dk|}=\frac{a+bi+cj+dk}{\sqrt{a^2+b^2+c^2+d^2}}$$ For example take $q=1+2i+3j+5k$. Then $$\operatorname{sgn}(1+2i+3j+5k)=\frac{1+2i+3j+5k}{|1+2i+3j+5k|}=\frac{1}{\sqrt{39}}+\frac{2}{\sqrt{39}}i+\frac{3}{\sqrt{39}}j+\frac{5}{\sqrt{39}}k$$ where $$\left(\frac{1}{\sqrt{39}}\right)^2+\left(\frac{2}{\sqrt{39}}\right)+\left(\frac{3}{\sqrt{39}}\right)^2+\left(\frac{5}{\sqrt{39}}\right)^2=1$$

Note in the complex case we have the modulus $|z|=\sqrt{z\cdot\bar{z}}=\sqrt{(a+bi)\cdot(a-bi)}=\sqrt{a^2+b^2}$, where $\bar{z}$ is the complex conjugate. In the quaternion case we have $|q|=\sqrt{q\cdot\bar{q}}=\sqrt{(a+bi+cj+dk)\cdot(a-bi-cj-dk)}=\sqrt{a^2+b^2+c^2+d^2}$, where $\bar{q}$ is the quaternion conjugate.

• Neither answer really takes on the "practical use" part of the question but this question is essentially "What is the practical use of orientation?" I don't have a great answer other than to say that orientation is helpful... in getting yourself orientated. Any take on the practical use part? +1 for your answer: obviously this page should include an answer which shows the steps in computing this. – Mason Jul 4 '18 at 21:16
• The utility of doing this is when constructing a basis in a vector space: it is normally most convenient to choose some orthonormal basis, where all vectors are normalised to unit length. – Daniel Buck Jul 4 '18 at 21:35
• Every basis can be orthogonal, i.e., such that for any two basis vectors $\mathbf{v_1}$, and $\mathbf{v_2}$ the dot product is $\mathbf{v_1}\cdot\mathbf{v_2}=0$. A basis of vectors is used to span either all or some subspace of a vector space such as $\Bbb{R}^n$. Look up the Gram-Schmidt orthogonalisation procedure for information on creating bases, and the topic of normalisation. – Daniel Buck Jul 4 '18 at 21:49