Mind-bending complex number questions

[JackSonSmith]

Could someone please work through these questions to show me how to do them, thank-you.

There are two separate questions attached as a word doc.

[wobblywobbly]

The first one is an annoying question to do, but basically you need to substitute z=x+yi, and turn the | | into sqrt form (remember || = magnitude). You would then need to move one of the square roots over to the other side, and square both sides. Keep simplifying until you can do no more, and you should still be stuck with one square root on one side, and whole terms on the other. Square both sides again and simplify to get your ellipse.

For the second one, consider
cis theta = cos theta + i sin theta
cis -theta = cos -theta + i sin -theta

Note cos theta = cos -theta and sin -theta = - sin theta so
cis -theta = cos theta - sin theta

[nerdmmb]

Could someone please work through these questions to show me how to do them, thank-you.

There are two separate questions attached as a word doc.

For your first question, sub z= x+ yi and square everything to get rid of the magnitudes.

(x+3)² + y² + (x-3)² + y² = 8²

x² + 6x + 9 + y² + x² - 6x + 9 = 8

2x² + 2y² = 64 - 18

2(x² + y²) = 46

Now divide both sides by 46 to make the right hand side equal to 1.

(x²)/23 + (y²)/23 = 1

And now that it is in the standard ellipse form- you can draw your ellipse with centre at the origin. I'm just a bit confused as to whether it's a horizontal or vertical ellipse because the a and b values are both the same.

[99.90 pls]

2(x² + y²) = 46

Now divide both sides by 46 to make the right hand side equal to 1.

(x²)/23 + (y²)/23 = 1

And now that it is in the standard ellipse form- you can draw your ellipse with centre at the origin. I'm just a bit confused as to whether it's a horizontal or vertical ellipse because the a and b values are both the same.

If a = b, then just leave it as x² + y² = 23
It's a circle with root 23 radius centred at (0, 0)

[nerdmmb]

If a = b, then just leave it as x² + y² = 23
It's a circle with root 23 radius centred at (0, 0)

OMG yes you're right.


LOL we were learning about ellipses yesterday and I think I was solving the qn with my mind in ellipse mode. Haha !

[nerdgasm]

For your first question, sub z= x+ yi and square everything to get rid of the magnitudes.

(x+3)² + y² + (x-3)² + y² = 8²

x² + 6x + 9 + y² + x² - 6x + 9 = 8

2x² + 2y² = 64 - 18

2(x² + y²) = 46

Now divide both sides by 46 to make the right hand side equal to 1.

(x²)/23 + (y²)/23 = 1

And now that it is in the standard ellipse form- you can draw your ellipse with centre at the origin. I'm just a bit confused as to whether it's a horizontal or vertical ellipse because the a and b values are both the same.

I would advise you to be wary of your algebra here. The starting equation we have is:

. Squaring it doesn't lead to your first line.

One way to check if your final answer is right, is to interpret as meaning, "the distance from plus the distance from is 8.
This allows you to quickly find points that lie on the ellipse, such as and , and also and .

It is relatively easy to see that these points do not satisfy . The answer that I get (following the method presented by wobblywobbly) is .