Complex Numbers- Square root 3i

[JC_003]

(1- sqrt 3i) ^3

I am having trouble solving the sqrt 3i part. I think I need to use de moivres theorem but I am unsure. If someone could push me in the right direction that would be a massive help. Thanks.

 

[Cyosis]

What's the question exactly? You want to write that complex number in its polar representation or as x+iy? Either way start with writing 1-\sqrt{(3i)} in the polar representation. Once you have that it is easy to compute (1-\sqrt{(3i)})^3.

 

[Дьявол]

I think he meant:

\sqrt[3]{1-i\sqrt{3}}

Use De Moivre's theorem to find all the solutions.

 

[JC_003]

The question is:

If z= 5 +4i, write the number z(|z|^2-(1- sqrt3i)^3) in the form a+bi.

I am able to sub everything in but I want to simplify the (1- sqrt3i)^3. The square root has thrown me off.

 

[Cyosis]

I take it the i is inside the square root? Try to write it in the form 3i=|z|e^{i\phi} then take the square root on both sides.

 

[JC_003]

Is that meant to the theta or phi? Silly question I know. I have in my notes a simalar formula for Eulers formula but that uses theta not phi... Another silly question I know but by changing the sq root 3i into a simple number going to give me the same result as getting the polar representation of the whole (1 - sqrt3i) ^3?

 

[Cyosis]

It doesn't matter whether it's called phi, theta or JC_003 it is just a variable which represents the angle between |z| and the positive real axis. You will get the same answer both ways, as it should. However the angle theta is pretty hard to find for 1-\sqrt{3i}. I suggest you write \sqrt{3i}=x+iy first and then continue from there.

 

[Elucidus]

If this is an exercise from a text, I'd be very carefull about whether the question concerns

(\sqrt{3})i​

or

\sqrt{(3i)}​

Huge difference.

The question is not clear enough for me to help effectively, but I suspect -8 shows up somewhere.

--Elucidus