# Thread: Plotting solutions on Argand diagram

1. ## Plotting solutions on Argand diagram

Plot the solutions of $\displaystyle z^2 = -2 + \sqrt{12}i$ on an Argand diagram

Attempt:

I found 2 solutions:

$\displaystyle z = \sqrt{2(-1 + i\sqrt{3})}$

$\displaystyle z = -\sqrt{2(-1 +i\sqrt{3})}$

but I am not sure what I am suppose to do next

2. Originally Posted by SyNtHeSiS
Plot the solutions of $\displaystyle z^2 = -2 + \sqrt{12}i$ on an Argand diagram

Attempt:

I found 2 solutions:

$\displaystyle z = \sqrt{2(-1 + i\sqrt{3})}$

$\displaystyle z = -\sqrt{2(-1 +i\sqrt{3})}$

but I am not sure what I am suppose to do next
You should find the solutions in polar form by using deMoivre's Theorem.

3. I got:

r = 4
$\displaystyle \theta = \frac{2\pi}{3} =\pi - \frac{\pi}{3}$ (since this angle lies in the 2nd quadrant)

$\displaystyle z^2 = 4(cos\frac{2\pi}{3} + isin\frac{2\pi}{3})$
$\displaystyle z^2 = 4(-cos\frac{2\pi}{3} + isin\frac{2\pi}{3})$ (since cos is negative in 2nd quadrant)

I am not sure what else to do

4. Originally Posted by SyNtHeSiS
Plot the solutions of $\displaystyle z^2 = -2 + \sqrt{12}i$ on an Argand diagram

Attempt:

I found 2 solutions:

$\displaystyle z = \sqrt{2(-1 + i\sqrt{3})}$

$\displaystyle z = -\sqrt{2(-1 +i\sqrt{3})}$

but I am not sure what I am suppose to do next
De Moivre's theorem is most likely the method of solution expected.

You have $\displaystyle z^2=4\left(Cos\frac{2{\pi}}{3}+iSin\frac{2{\pi}}{3 }\right)$

Your second expression is incorrect and unnecessary, since $\displaystyle z^2$ is a single point on the complex plane.

However, it does have 2 square roots.

De Moivre's Theorem gives $\displaystyle z=|z|^{\frac{1}{2}}\left(Cos\beta+iSin\beta\right) ^{\frac{1}{2}}$

$\displaystyle =\sqrt{4}\left(Cos\frac{\beta}{2}+iSin\frac{\beta} {2}\right)$

You will have to use the 2 values of $\displaystyle \beta$,...... $\displaystyle \frac{2{\pi}}{3},\;\;2{\pi}+\frac{2{\pi}}{3}$

Alternatively,

$\displaystyle z=|z|e^{i\theta}\Rightarrow\ z^2=|z|^2e^{i2\theta}$

Using Pythagoras' theorem, the length of $\displaystyle z^2$ is $\displaystyle 4$

Therefore the length of the square root is 2.

The argument of $\displaystyle z^2$ is $\displaystyle arctan\left(-\frac{\sqrt{12}}{\sqrt{4}}\right)=arctan(-\sqrt{3})={\pi}-arctan{\sqrt{3}}=\frac{2{\pi}}{3}$

To find solutions for $\displaystyle z$ from $\displaystyle 0^o\rightarrow\ 2{\pi}$, realise that in squaring $\displaystyle z$, the double angle of $\displaystyle z^2$ is $\displaystyle <4{\pi}$

We therefore also take a 2nd angle for $\displaystyle z$, the angle $\displaystyle 2{\pi}+\frac{2{\pi}}{3}$

These angles are $\displaystyle 2\theta\Rightarrow\ \theta=\frac{{\pi}}{3},\;\;{\pi}+\frac{{\pi}}{3}$

5. Hello, SyNtHeSiS!

There is an algebraic solution . . .

$\displaystyle \text{Plot the solutions of }\,z^2 \:=\: -2 + \sqrt{12}i\,\text{ on an Argand diagram.}$

$\displaystyle \text{Let: }\:z \:=\: a + bi\:\text{ where }a\text{ and }b\text{ are real number.}$

$\displaystyle \text{We have: }\:(a+bi)^2 \:=\:-2 + 2\sqrt{3}\,i$

. . .$\displaystyle (a^2 - b^2) + 2abi \:=\:-2 + 2\sqrt{3}\,i$

Equate real and imaginary components: .$\displaystyle \begin{Bmatrix}a^2-b^2 &=& -2 & [1] \\ 2ab &=& 2\sqrt{3} & [2] \end{Bmatrix}$

From [2]: .$\displaystyle 2ab \:=\:2\sqrt{3} \quad\Rightarrow\quad b \:=\:\frac{\sqrt{3}}{a}$ .[3]

Substitute into [1]: .$\displaystyle a^2 - \left(\frac{\sqrt{3}}{a}\right)^2 \:=\:-2 \quad\Rightarrow\quad a^2 - \frac{3}{a^2} \:=\:-2$

. . . . . . . $\displaystyle a^4 + 2a^2 - 3 \:=\:0 \quad\Rightarrow\quad (a^2-1)(a^2+3) \:=\:0$

. . $\displaystyle \begin{array}{cccccccccc}a^2 - 1 \:=\:0 & \Rightarrow & a^2 \:=\:1 & \Rightarrow & a \:=\:\pm1 \\ a^2+3 \:=\:0 & \Rightarrow& a^2 \:=\:\text{-}3 & \Rightarrow & a \:=\:\pm\sqrt{3}\,i \end{array}$

Since $\displaystyle \,a$ is real, $\displaystyle a \:=\:\pm1$

Substitute into [3]: .$\displaystyle b \:=\:\frac{\sqrt{3}}{\pm1} \quad\Rightarrow\quad b \:=\:\pm\sqrt{3}$

Therefore: .$\displaystyle z \;=\;\pm1 \pm\sqrt{3}\,i \;=\;\pm(1 + \sqrt{3}\,i)$

Now plot those two points . . .

6. Originally Posted by Archie Meade
You will have to use the 2 values of $\displaystyle \beta$,...... $\displaystyle \frac{2{\pi}}{3},\;\;2{\pi}+\frac{2{\pi}}{3}$
Why does B have 2 values why not only $\displaystyle \frac{2{\pi}}{3}$?

7. Originally Posted by SyNtHeSiS
Why does B have 2 values why not only $\displaystyle \frac{2{\pi}}{3}$?
From an algebraic perspective, $\displaystyle z^2$ has two square roots, $\displaystyle z$ and $\displaystyle -z$

so the square root of $\displaystyle z^2$ at angle $\displaystyle \frac{{\pi}}{3}$ is $\displaystyle 1+\sqrt{3}i$

The negative square root is $\displaystyle -1-\sqrt{3}i$

These complex numbers differ by $\displaystyle {\pi}$ radians.

The reason we take 2 angles $\displaystyle \beta$ is because in the complex world, the angles

$\displaystyle \frac{2{\pi}}{3}$ and $\displaystyle 2{\pi}+\frac{2{\pi}}{3}$ are the same, as they reference the exact same position.

When a complex number is squared, it's angle doubles,
so a complex number with an angle between $\displaystyle {\pi}$ and $\displaystyle 2{\pi}$
when squared has an angle between $\displaystyle 2{\pi}$ and $\displaystyle 4{\pi}$.