Is there a number that ${x^2=i}$ where ${i^2=-1}$

Question

Imagine that there is a number ${x^2=i}$ where ${i^2=-1}$ where should it be as it is not in ${C}$ , and how to draw it , and where ?

The easiest way to think about these things is in terms of polar coordinates. But yes, there are two complex numbers that make $x^2 = i$ true. — Braindead, Jan 04 '14 at 19:57
Check out Fundamental Theorem of Algebra which give existence answer to question about root of polynomial. — Gina, Jan 04 '14 at 20:09

score 5 · Answer 1 · answered Jan 04 '14 at 19:56

There is absolutely such an $x$ (two, in fact), and they both do lie in $\mathbb{C}$.

The nice thing about squares in complex numbers is that they work in a very predictable way: they double the argument and square the modulus.

The complex number $i$ has modulus $1$ and argument $\frac{\pi}{2}$. This suggests immediately that the number which has modulus $1$ and argument $\frac{\pi}{4}$ -- namely, $\frac{1}{\sqrt{2}}+i\frac{1}{\sqrt{2}}$ -- is a square root of $i$, for instance.

Note that you should have expected this to have solutions in $\mathbb{C}$, as $\mathbb{C}$ is algebraically closed: that is, all polynomials with coefficients in $\mathbb{C}$ factor completely in $\mathbb{C}$. Such a number $x$ is a root of $x^2-i=0$.

score 5 · Answer 2 · answered Jan 04 '14 at 20:00

Recall that complex numbers can be written in polar form:

$$a+bi=\sqrt{a^{2}+b^{2}}e^{i\varphi}$$

Where $\varphi=\arg(a+bi)$, in this case we have $\sqrt{a^{2}+b^{2}}=1$ and $\varphi=\arg(i)=\frac{\pi}{2}$. We therefore have:

$$i=e^{\frac{i\pi}{2}}\implies x=\pm(e^{\frac{i\pi}{2}})^{\frac{1}{2}}=\pm e^{\frac{i\pi}{4}}=\pm\frac{1}{\sqrt{2}}(1+ i)$$

score 4 · Answer 3 · answered Jan 04 '14 at 20:02

4

Since $$\Large i=e^{i\frac \pi 2}$$ then $$\Large x=\pm e^{i\frac \pi 4}\Rightarrow x^2=i$$

answered Jan 04 '14 at 20:02

The bigger words the better we can read them. :+) – Mikasa Jan 05 '14 at 07:46

score 3 · Answer 4 · edited Jan 04 '14 at 20:01

3

$(1+i)^{2} = 1 + 2i - 1 = 2i$. Divide by $2$ to get $i$.

edited Jan 04 '14 at 20:01

Salech Alhasov

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answered Jan 04 '14 at 19:56

neelp

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score 3 · Accepted Answer · answered Jan 04 '14 at 20:18

Lets, instead of $x^2=i$, we write $z^2=i$, where $z=u+iv$, here $u,v\in\mathbb{R}$.

$$z^2=(u+iv)^2=i$$

We know that, $(u+iv)^2=u^2+2ivu-v^2$

Therefore,

$$z^2=(u+iv)^2=u^2+2ivu-v^2=i$$

Solving

$$u^2+2ivu-v^2=i$$

We will have $u^2-v^2$, the "real part", is must be zero, since in the right side of the equation there is no "real part". And $2uv$, the "complex part" must be $1$, since in the right side of the equation there is $i$.

Therefore, the task is finding those $u,v$, and that we doing by solving the following system of equations:

$$\left\{\begin{matrix} u^2-v^2=0\\ 2uv=1 \end{matrix}\right.$$

Try to finish this off.

score 1 · Answer 6 · answered Jan 04 '14 at 19:56

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Try squaring $(1 + i)/\sqrt{2}.$ I'll bet you can easily find another!

answered Jan 04 '14 at 19:56

ncmathsadist

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Is there a number that ${x^2=i}$ where ${i^2=-1}$

6 Answers6