What is i to the power of 1/2

Question

I am new to complex numbers and wondered, what is $i^{(1/2)}$ and how do I actually create a complex number, so that $i$ has a different power than $1$ in tuple form, like $(a, b) = a + b*i^2$? Is it just $(a, bi)$?

Answer 1

In the complex plane, $i$ is on the unit circle at an angle of $\frac{\pi}{2}$. Thus it has two square roots, which are obtained by taking half the angle and accounting for wraparound by $2\pi$.

Thus, the most obvious square root is the point on the unit circle with angle $\frac{\pi}{4}$. The less obvious square root is obtained by replacing $\frac{\pi}{2}$ with the equivalent angle $2\pi+\frac{\pi}{2}$, and taking half of it results in $\frac{5\pi}{4}$. Note that the second square root is diametrically opposite from the first square root on the unit circle.

To obtain the coordinates (real and imaginary parts) of these two square roots, you can use the fact that a point at angle $\theta$ on the unit circle has coordinates $(\cos \theta,\sin\theta)$ and substituting the values $\theta=\frac{\pi}{4}$ and $\theta=\frac{5\pi}{4}$ into this.

Answer 2

You should learn about the polar form of complex numbers: https://www.varsitytutors.com/hotmath/hotmath_help/topics/polar-form-of-a-complex-number

Any complex $z$ can be written as $z=Re^{i\theta}$, where $R$ is the "magnitude," $R=|z|$, and $\theta$ is the angle you have to turn to get to $z$, starting from the real axis.

You should think of polar form as cutting the complex plane into circles centered at the origin; $R$ tells you what circle $z$ is on, and $\theta$ tells you where $z$ is on that circle.

Now it is an easy fact that if $z=Re^{i\theta}$, then $z^n = R^ne^{ni\theta}$. Thus to compute a square root of $i$ we should first write $i$ in polar form, $$ i=e^{i\pi/2}, $$ and then we easily get two square roots $$ \pm e^{i\pi/4}. $$ You can then translate this back if you want to rectangular form to get the roots $$ \pm \frac{1+i}{\sqrt{2}} $$

Answer 3

Think it out.

If $i^{\frac 12} = a + bi$ for real $a, b$ then

$i=(i^{\frac 12})^2 = (a + bi)^2$

$i = a^2 + 2ab i + b^2i^2$

$a^2 + 2abi + b^2(-1) = i$

$(a^2 - b^2) + 2ab i = 0 + 1*i$

So the real components of $i$ are $a^2 -b^2 =0$ and $2ab =1$.

Can you finish up?

So $a^2 = b^2$ and $|a| = |b|$ and $a =\pm b$ so

$2ab = 1$

$\pm 2b^2 = 1$ so

$\pm b^2 =\frac 12$. As $b$ is real that means $b^2$ is positivee so

$b^2 =\frac 12$ and $a = b$

So $b = \pm \frac 1{\sqrt 2} = \pm \frac {\sqrt 2}{2}$.

And $a = \pm \frac {\sqrt 2}{2}$.

So $i^{\frac 12} = \pm (\frac {\sqrt 2}{2} + \frac {\sqrt 2}{2}i)$

.....

Verify $[\pm(\frac {\sqrt 2}{2} + \frac {\sqrt 2}{2}i)]^2 =$

$(\frac {\sqrt 2}{2})^2 + 2\frac {\sqrt 2}{2}\frac {\sqrt 2}{2}i + (\frac {\sqrt 2}{2}i)^2 =$

$\frac 24 + 2\frac 24i + \frac 24i^2 = $

$\frac 24 + i - \frac 24 = i$.

Ta-da, as it were.

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how do I actually create a complex number, so that i has a different power than 1 in tuple form

This actually doesn't matter. If you ever need to express $i^k$ for $k\in \mathbb C$ it will turn out (trust me on this) that there are real numbers $w,v$ so that $i^k = w + vi$.

So it doesn't matter.

If $k \in \mathbb Z$ then $k = 4m + r$ for some integer $m$ and $r = 0,1,2,3$. Note $i^1=i$ and $i^2 = -1$ and $i^3= i^2*i = -i$ and $i^4 = (i^2)^2 = (-1)^2 = 1 =i^0$. So $i^k = i^{4m+r} = (i^4)^mi^r = 1^m*i^r = i^r =1,i, -1,-i$ depending if $r = 0,1,2,3$.

If $k = \frac mn\in \mathbb Q$ we can figure that if $i^k = w+vi$ then $(i^k)^n = i^m = (w+vi)^n$. $(w+vi)^n$ we expand to a polynomial but all the power of $i$s will reduce to elther $\pm 1$ or $\pm i$ and the whole thing becomse some $c + di= i^m$ where $c,d$ are combinations of $w,v$ which can be solved. (One is $\pm 1$ and the other is $0$).

For $i^k$ where $k$ is irrational or even complex are a bit trickier but even then....

Once you learn polar coordinates where you think of $a_bi$ as a point $(a,b)$ in a plane we can express $z = a+ bi$ terms of an angle and distance from the origin.

We learn a neat trick from trigonometry: If $z = a+bi$ translates to distance $r$ and angle $\theta$and $w = c+di$ translates to distance $s$ and angle $\eta$ then $z*w$ translates to distance $rs$ and angle $\theta + \eta$.

Forthurmore the distance $r$ is the distance of $(a,b)$ from the origin $(0,0)$ is $r=\sqrt{a^2 + b^2}$ and $a = r\cos \theta$ and $b=r\sin \theta$.

But that is all food for thought down the line.

Answer 4

Note that $$ i=cos(\pi/2)+i\sin (\pi/2)$$

You are looking for complex numbers $$z=\cos (\theta)+i\sin(\theta)$$ such that $z^2=i$

That is $$\cos(2\theta)+i\sin(2\theta)=cos(\pi/2)+i\sin (\pi/2)$$

Clearly $\theta = \pi/4$ and $ \theta = 5\pi/4$ satisfy the equation.

Thus you have two answers. $$ z=\frac {\sqrt 2}{2}(1+i)$$ and $$z=\frac {-\sqrt 2}{2}(1+i)$$

Answer 5

In reverse order:

$$a+bi^2 = a + b(-1) = (a-b) + 0i = (a-b, 0)$$

Since the tuple form comes from the standard form of the complex number, there never is a higher power on the $i$ than $1.$

As for powers of $i$... We can find two possibilities for $i^{1/2}$ just by solving $$(a+bi)^2 = i \\ (a^2-b^2) + (2ab)i = i \\ \begin{cases}a^2-b^2 = 0 \\ 2ab = 1\end{cases} \\ a = b = \pm \frac{\sqrt{2}}{2} \\ a+bi = \pm\left(\frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2}i\right)$$

A word of warning: While the powers of positive real numbers can be defined as a single number, you will almost certainly run into problems if you want a consistent definition of powers of complex numbers. In the most general case, the power $(a+bi)^r$ is evaluated as $e^{r\ln(a+bi)},$ which means in order to find a power of a complex number, we first need to define the natural logarithm, which requires a lot of care.