Complex Number

1. Feb 21, 2010

icystrike

1. The problem statement, all variables and given/known data
Express the complex number in exp. form
$$-\frac{1}{2}(1+i\sqrt{3})$$

Solve the following eqn:
$$(w+2)^{4}=-\frac{1}{2}(1+i\sqrt{3})$$

2. Relevant equations

3. The attempt at a solution
$$e^{i\frac{\pi}{3}}$$

$$w+2=e^{(\frac{\frac{\pi}{3}+2k\pi}{4})i} =e^{(\frac{\pi}{12}+\frac{\pi}{2}k)i}$$

Such that k=0,1,2,3

2. Feb 21, 2010

tiny-tim

Hi icystrike!

So, to get minus that, multiply by … ?

For the second answer, the ekπi/2 can be simplified a little, for example by using a "±" (and of course, put the 2 on the other side).

3. Feb 21, 2010

icystrike

multiply by -1 . so it means that we have to keep a look out to or not multiply the ex. form by -1 right?

4. Feb 21, 2010

tiny-tim

uhh?

Hint: e? = -1 ?

5. Feb 21, 2010

icystrike

multiply by $$e^{i\pi}$$ thus combine the power by law of indices

6. Feb 21, 2010

tiny-tim

Yup! … so instead of eiπ/3, it's … ?

7. Feb 21, 2010

icystrike

$$e^{\frac{4\pi}{3}}$$

yea?

8. Feb 22, 2010

tiny-tim

(just got up :zzz: …)

yea!

9. Feb 22, 2010

icystrike

Hi tiny-tim! Can you help me with this question?

Explain why the equation $$(z+2i)^{6}=z^{6}$$ has five roots.

I thought it should be 6 roots?

10. Feb 22, 2010

tiny-tim

Hi icystrike!

erm … it's only a fifth-order equation!

11. Feb 22, 2010

icystrike

hmm.. how do you tell? always thought that if we have power 6 , it will be 6 roots.

12. Feb 22, 2010

tiny-tim

You don't like algebra, do you?

Expand the LHS, and subtract the RHS … what happens?

13. Feb 22, 2010