# Can a complex number be written Cartesian form in terms of i to a power other than 1? And if so...

## I came across the question: ${z}_{1} = - 2 - 2 \sqrt{3 i}$ ${z}_{2} = 3 \sqrt{3} + 3 i$ Find ${z}_{3} = {z}_{1} {z}_{2}$ in Cartesian form. But I keep ending up with multiple coefficients in terms of i, but each to a different power, such as $- 6 \sqrt{3} + 6 i \sqrt{3 i} - 18 \sqrt{i} - 6 i$. How can this be written in Cartesian form properly, or did I do something wrong to reach this?

Jul 24, 2017

The square roots of $\text{i}$ are $\pm \frac{1}{\sqrt{2}} \left(1 + i\right)$. So $\sqrt{\text{i") = 1/sqrt(2)(1+"i}}$.

#### Explanation:

Yes, this number can be written in Cartesian form. The problem at the moment is that you have a $\sqrt{\text{i}}$.

Your problem seems to be with taking the square root of a complex number. There are two ways to do this: one with just knowledge of basic algebra and one (far more useful) with knowledge of Euler's identity.

Way 1: Basic algebra.

Let $z = \sqrt{\text{i}}$ and $z = a + b \text{i}$, where $a , b \in \mathbb{R}$.

Then,

${\left(a + b \text{i}\right)}^{2} = i$,
${a}^{2} + 2 a b \text{i"-b^2="i}$,
${a}^{2} - {b}^{2} + \left(2 a b\right) \text{i"=0+"i}$.

Equating coeffecients ${a}^{2} - {b}^{2} = 0$ and $2 a b = 1$.

${a}^{2} - {b}^{2} = 0 \implies a = \pm b$.

Then $2 {a}^{2} = \pm 1$. As $a \in \mathbb{R}$, $a = b = \frac{1}{\sqrt{2}}$. We conclude that the square roots of $\text{i}$ are $\pm \frac{1}{\sqrt{2}} \left(1 + i\right)$.

We can check this by squaring $\frac{1}{\sqrt{2}} \left(1 + \text{i}\right)$.

Way 2: Eulers identity

Let $z$ be the complex number you want to raise to an exponent. A complex number can be written as $z = r {e}^{\text{i} \theta}$.

$r {e}^{\text{i"theta)=r(cos(theta)+"i} \sin \left(\theta\right)}$ by eulers identity.

The sine and cosine functions have periodicity $2 \pi$. So any complex number $z$ can be written as $z = r {e}^{\text{i} \theta + 2 k \pi}$ with $k \in \mathbb{Z}$.

Then,

${z}^{n} = {r}^{n} {e}^{n \left(\text{i} \left(\theta + 2 k \pi\right)\right)}$ for $n \in \mathbb{R}$.

So, for this specific example, let $z = i$. Then $\left\mid z \right\mid = r = \sqrt{{1}^{2}} = 1$.
The number $\text{i}$ makes a anticlockwise angle of $\frac{\pi}{2}$ from the positive $x$ axis.

Then we conclude z="i"=e^("i"pi(1/2+2k)).

${z}^{\frac{1}{2}} = {e}^{\text{i} \frac{\pi}{2} \left(\frac{1}{2} + 2 k\right)}$.

Then choose $k$ such that we receive all the arguments between $- \pi$ and $\pi$.

$k = 0 \implies {z}^{\frac{1}{2}} = {e}^{\text{i} \frac{\pi}{4}}$,
$k = - 1 \implies {z}^{\frac{1}{2}} = {e}^{- \text{i} \frac{3 \pi}{4}}$.

When we raise to a power $\frac{1}{n}$ $n \in \mathbb{Z}$ we expect $n$ solutions for $z$. In this case $n = 2$ and we have found the two solutions. By writing e^("i"theta)=cos(theta)+"i"sin(theta) you will see that these two solutions for ${z}^{\frac{1}{2}}$ give the square root of $\text{i}$ as $\pm \frac{1}{\sqrt{2}} \left(1 + \text{i}\right)$.

I hope you can see how this method is more applicable to finding the roots of complex numbers more generally (and demonstrates that they can always be written in Cartesian form).

I trust you can use $\sqrt{3 \text{i")=sqrt("3")sqrt("i")=sqrt("3")/sqrt("2")(1+"i}}$ to finish the question you asked!