# Complex Numbers- Square root 3i

1. Aug 12, 2009

### 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.

2. Aug 12, 2009

### 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 $(1-\sqrt{(3i)})^3$.

Last edited: Aug 12, 2009
3. Aug 12, 2009

### Дьявол

I think he meant:

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

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

4. Aug 12, 2009

### 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.

5. Aug 12, 2009

### 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.

6. Aug 12, 2009

### 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?

7. Aug 12, 2009

### 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.

8. Aug 12, 2009

### 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