# Finding z1 , z2 of complex numer

1. May 31, 2014

### kelvin macks

1. The problem statement, all variables and given/known data

in order to get z1 and z2 , i tried to express z^2 into polar form, but z is to the power of 2, i'm not sure whether it can be epressed in polar form of not. by the way , here's my working. how should i proceed? i dont think my ans is correct.

2. Relevant equations

3. The attempt at a solution

#### Attached Files:

• ###### popp.jpg
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2. May 31, 2014

### ehild

I can not read your handwriting. The polar form can be used, but the solution is much simpler if you start writing z=u+iv and solving for u, v. Try.

ehild

3. May 31, 2014

### mafagafo

I can only understand the numbers.

4. May 31, 2014

### Ray Vickson

5. May 31, 2014

### kelvin macks

do u mean i should let z = a+bi , then i sub z = a+bi into z^2 ? by doing so, you assume that z1 and z2 , one of it must be conjugate to the others. am i right? here's the sample ans, why the author let z = a-bi , and not z = a+bi ?

#### Attached Files:

• ###### DSC_0014~4[1].jpg
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18.5 KB
Views:
38
Last edited: May 31, 2014
6. May 31, 2014

### mafagafo

I am not ehild, but I really do think that this is what he meant.

7. May 31, 2014

### kelvin macks

can it be done using plar form? if can, can you please show me how do u express z^2 in polar form, i have done it in the very first post, but i dont think it's correct. my ans is weird.

8. May 31, 2014

### mafagafo

can it be done using plar form?
Yeah.

if can, can you please show me how do u express z^2 in polar form
I can, but will not. I like the policy of not giving full answers. For anything.

Hint: do you know De Moivre's formula?
if not, then https://en.wikipedia.org/wiki/De_Moivre's_formula is a good place to start.

There are some other places for it too. The solution you attached in #5 is rock-solid.

Edit: share your attempt at of expressing z^2 using De Moivre's so we can check it for you. Use LaTeX or ASCII characters, you handwriting is... complicated.

9. Jun 1, 2014

### ehild

No, z1 and z2 are not conjugate pairs, but one is the negative of the other. You can write the complex z as $z=\pm \sqrt{1-2\sqrt2 i}$

As the imaginary part of z2 is negative, it is convenient to consider the imaginary part of z also negative. It does not matter. So z = a-ib, you square it, and compare with 1-2√2 i, so as the real parts are equal and the imaginary parts are also equal. Solve the system of equation for a, b.
Note that both a and b are real.

You can use polar form, too, but it is tedious. You convert 1-2√2 i to polar form numerically, determine the second roots, and convert back. You round at each step, losing accuracy.
I suspect you did not use the correct formula for the n-th root of a complex number. Check it.

ehild

Last edited: Jun 1, 2014