# Plotting solutions on Argand diagram

• October 20th 2010, 04:05 AM
SyNtHeSiS
Plotting solutions on Argand diagram
Plot the solutions of $z^2 = -2 + \sqrt{12}i$ on an Argand diagram

Attempt:

I found 2 solutions:

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

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

but I am not sure what I am suppose to do next
• October 20th 2010, 04:13 AM
mr fantastic
Quote:

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

Attempt:

I found 2 solutions:

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

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

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.
• October 20th 2010, 08:47 AM
SyNtHeSiS
I got:

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

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

I am not sure what else to do
• October 20th 2010, 02:58 PM
Quote:

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

Attempt:

I found 2 solutions:

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

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

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

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

However, it does have 2 square roots.

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

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

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

Alternatively,

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

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

Therefore the length of the square root is 2.

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

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

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

These angles are $2\theta\Rightarrow\ \theta=\frac{{\pi}}{3},\;\;{\pi}+\frac{{\pi}}{3}$
• October 20th 2010, 03:57 PM
Soroban
Hello, SyNtHeSiS!

There is an algebraic solution . . .

Quote:

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

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

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

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

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

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

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

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

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

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

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

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

Now plot those two points . . .
• October 23rd 2010, 04:44 AM
SyNtHeSiS
Quote:

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

Why does B have 2 values why not only $\frac{2{\pi}}{3}$?
• October 23rd 2010, 12:05 PM
Quote:

Originally Posted by SyNtHeSiS
Why does B have 2 values why not only $\frac{2{\pi}}{3}$?

From an algebraic perspective, $z^2$ has two square roots, $z$ and $-z$

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

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

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

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

$\frac{2{\pi}}{3}$ and $2{\pi}+\frac{2{\pi}}{3}$ 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 ${\pi}$ and $2{\pi}$
when squared has an angle between $2{\pi}$ and $4{\pi}$.