Solving a Complex Equation: Express as e^(iθ)

[icystrike]

Homework Statement

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})

Homework Equations

The Attempt at a Solution

e^{i\frac{\pi}{3}}

w+2=e^{(\frac{\frac{\pi}{3}+2k\pi}{4})i}<br /> =e^{(\frac{\pi}{12}+\frac{\pi}{2}k)i}<br />

Such that k=0,1,2,3

 

[tiny-tim]

Hi icystrike!

Your first answer, e^πi/3, would be the correct answer for +1/2 (1 + i√3).

So, to get minus that, multiply by … ?

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

 

[icystrike]

Hi icystrike!

Your first answer, e^πi/3, would be the correct answer for +1/2 (1 + i√3).

So, to get minus that, multiply by … ?

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

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

 

[tiny-tim]

uhh?

Hint: e^? = -1 ? ​

 

[icystrike]

uhh?

Hint: e^? = -1 ? ​

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

 

[tiny-tim]

Yup! … so instead of e^iπ/3, it's … ?

 

[icystrike]

Yup! … so instead of e^iπ/3, it's … ?

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

yea?

 

[tiny-tim]

(just got up :zzz: …)

yea!

 

[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?

 

[tiny-tim]

Hi icystrike!

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

 

[icystrike]

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

 

[tiny-tim]

You don't like algebra, do you?

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

 

[icystrike]

Oh! the term z^{6} will vanish. how could i neglect that?
Once again a big thanks to you!
btw, do you mind to check out on this thread of mine?
https://www.physicsforums.com/showthread.php?t=380399