Plot the solutions of on an Argand diagram

Attempt:

I found 2 solutions:

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

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- Oct 20th 2010, 05:05 AMSyNtHeSiSPlotting solutions on Argand diagram
Plot the solutions of on an Argand diagram

Attempt:

I found 2 solutions:

but I am not sure what I am suppose to do next - Oct 20th 2010, 05:13 AMmr fantastic
- Oct 20th 2010, 09:47 AMSyNtHeSiS
I got:

r = 4

(since this angle lies in the 2nd quadrant)

(since cos is negative in 2nd quadrant)

I am not sure what else to do - Oct 20th 2010, 03:58 PMArchie Meade
De Moivre's theorem is most likely the method of solution expected.

You have

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

However, it does have 2 square roots.

De Moivre's Theorem gives

You will have to use the 2 values of ,......

Alternatively,

Using Pythagoras' theorem, the length of is

Therefore the length of the square root is 2.

The argument of is

To find solutions for from , realise that in squaring , the double angle of is

We therefore also take a 2nd angle for , the angle

These angles are - Oct 20th 2010, 04:57 PMSoroban
Hello, SyNtHeSiS!

There is an algebraic solution . . .

Quote:

. . .

Equate real and imaginary components: .

From [2]: . .[3]

Substitute into [1]: .

. . . . . . .

. .

Since is real,

Substitute into [3]: .

Therefore: .

Now plot those two points . . .

- Oct 23rd 2010, 05:44 AMSyNtHeSiS
- Oct 23rd 2010, 01:05 PMArchie Meade
From an algebraic perspective, has two square roots, and

so the square root of at angle is

The negative square root is

These complex numbers differ by radians.

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

and 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 and

when squared has an angle between and .