I have two different complex numbers. Say $a_1+b_1i$ and $a_2+b_2i$. I take the square root of these two complex numbers. Let $ \sqrt(a_1+b_1i) = c_1+d_1i $ and $ \sqrt(a_2+b_2i) = c_2+d_2i $. I can solve for $c_1~\&~d_1$ in terms of $a_1~\&~b_1$ (respectively $c_2~\&~d_2$ in terms of $a_2~\&~b_2$). I want $c_1=c_2$. I can solve algebraically under which the condition $c_1=c_2$ holds in terms of $(a_1,b_1,a_2,b_2)$. How do I interpret the condition $c_1=c_2$ geometrically in terms of $a_1,b_1,a_2,b_2$? I wish to understand the above condition in a geometrical way with reference to the original complex numbers.


1 Answer 1


First of all, "the" square root of a complex number is not well defined, since a complexe number has two opposite square roots. However, is does not matter much.

For a complex point $z = c + id$, you have $z^2 = (c^2-d^2) + 2icd$. So taking $c$ as a constant, $z^2$ lies on the parabola $P_c$ of equation $y^2 + 4c^2x-4c^4=0$ (it is a parabola "turned" to the left, except for $c=0$ in which case the set to look at is the half-line $\mathbf{R}^-$, which is a kind of degenerated and flat parabola). The family $(P_c)_{c \geq 0}$ forms a partition of the plane and two complex numbers have "square roots" with the same real part if and only if they lie on the same $P_c$.

Another way to say it it that the complex function $z \mapsto z^2$ maps the vertical lines into parabolas (except for the vertical axis). In general, complex functions give rise to interesting pictures. You should have a look at conformal mappings.

  • $\begingroup$ After a quick research, I found this nice illustration. $\endgroup$
    – BrL
    May 11, 2016 at 20:22
  • $\begingroup$ Thanks. I'm looking on conformal mappings as well. Thanks for the cool link. $\endgroup$
    – Jerry
    Aug 8, 2016 at 4:56

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