# Thread: [SOLVED] A complex number

1. ## [SOLVED] A complex number

Find all the complex numbers such that $\displaystyle z^2=1-i$.
My attempt : I wrote $\displaystyle z=(a+ib)$ and with a little bit of some algebra I reached that $\displaystyle ab=-\frac{1}{2}$ which mean that if $\displaystyle ab=-\frac{1}{2}$ then $\displaystyle z^2=1-i$ is satisfied. So the solution is all z such that $\displaystyle \forall$ a,b $\displaystyle \in \mathbb{R}$ / $\displaystyle ab=-\frac{1}{2}$ with $\displaystyle z=a+ib$. I'm sure there is a nicer way to write the answer. Can you check out my answer and write it in a better way if it is right? Thanks in advance.

2. You are correct that
$\displaystyle 2ab=-1$

but remember to equate the real parts as well as the imaginary parts

so

$\displaystyle (a+ib)^2=1-i$

$\displaystyle a^2+2abi-b^2=1-i$

$\displaystyle (a^2-b^2)+2abi=1-i$

equating real parts as well gives us $\displaystyle a^2-b^2=1$

now solve the two equations simultaneously to find the answer

as an aside you can always check your answer by plugging a few numbers in, if we try a=1 b=-0.5 which satisfy your condition we have
$\displaystyle (1-\frac{1}{2}i)^2=1-\frac{1}{4}-1i=\frac{3}{4}-1i\neq 1-i$

therefore it cannot be correct on its own

3. but remember to equate the real parts as well as the imaginary parts

so

equating real parts as well gives us
I didn't forget to do that, the problem is that I didn't reached anything beautiful. Like $\displaystyle b=+$ or $\displaystyle -\sqrt{a^2-1}$, or $\displaystyle b^4-b^2=\frac{1}{4}$, or $\displaystyle 4b^2(b^2+1)=1$ and so on.
I'd like to know the answer and then try to reach it.
I've a similar problem to solve after this so I'll try it alone.

4. Why not just find the two square roots of $\displaystyle 1-i$?

5. Originally Posted by arbolis
I didn't forget to do that, the problem is that I didn't reached anything beautiful. Like $\displaystyle b=+$ or $\displaystyle -\sqrt{a^2-1}$, or $\displaystyle b^4-b^2=\frac{1}{4}$, or $\displaystyle 4b^2(b^2+1)=1$ and so on.
I'd like to know the answer and then try to reach it.
I've a similar problem to solve after this so I'll try it alone.
you are finding the two square roots of 1 - i

the answers are $\displaystyle a = \pm \sqrt{\frac {1 + \sqrt{2}}2}$ and $\displaystyle b = - \frac 1{2a}$ ........and of course, you plug in what a is (can you get to that? )

you could also try converting to polar form, it might be the same, or more work, but not much different in terms of difficulty

6. I've reached $\displaystyle b=\pm \sqrt{\frac{1}{\pm 2\sqrt 2 +2}}$ so you can imagine to what $\displaystyle a$ is equal. But $\displaystyle b$ is a real number, so I can simply it to $\displaystyle \pm \sqrt{\frac{1}{2\sqrt 2 +2}}$. I feel I made an error somewhere.
I've started with $\displaystyle a^2-b^2=-2ab$.

7. In polar/exponential form $\displaystyle 1 - i = \sqrt 2 \left( {\cos \left( {\frac{{ - \pi }}{4}} \right) + i\sin \left( {\frac{{ - \pi }}{4}} \right)} \right) = \sqrt 2 e^{i\left( {\frac{{ - \pi }}{4}} \right)}$
The roots are $\displaystyle z_1 = \sqrt[4]{2}\left( {\cos \left( {\frac{{ - \pi }} {8}} \right) + i\sin \left( {\frac{{ - \pi }}{8}} \right)} \right)\;\& \;z_2 = \sqrt[4]{2}\left( {\cos \left( {\frac{{7\pi }}{8}} \right) + i\sin \left( {\frac{{7\pi }}{8}} \right)} \right)$

8. Originally Posted by Plato
In polar/exponential form $\displaystyle 1 - i = \sqrt 2 \left( {\cos \left( {\frac{{ - \pi }}{4}} \right) + i\sin \left( {\frac{{ - \pi }}{4}} \right)} \right) = \sqrt 2 e^{i\left( {\frac{{ - \pi }}{4}} \right)}$
The roots are $\displaystyle z_1 = \sqrt[4]{2}\left( {\cos \left( {\frac{{ - \pi }} {8}} \right) + i\sin \left( {\frac{{ - \pi }}{8}} \right)} \right)\;\& \;z_2 = \sqrt[4]{2}\left( {\cos \left( {\frac{{7\pi }}{8}} \right) + i\sin \left( {\frac{{7\pi }}{8}} \right)} \right)$
Ah that's nice, I think I will do that in my exam even if I should simplify $\displaystyle -\frac{\pi}{8}$ and $\displaystyle \frac{7\pi}{8}$. I think my teacher will give me as much credit as if I solve it without using polar form.

9. Originally Posted by arbolis
I've reached $\displaystyle b=\pm \sqrt{\frac{1}{\pm 2\sqrt 2 +2}}$ so you can imagine to what $\displaystyle a$ is equal. But $\displaystyle b$ is a real number, so I can simply it to $\displaystyle \pm \sqrt{\frac{1}{2\sqrt 2 +2}}$. I feel I made an error somewhere.
I've started with $\displaystyle a^2-b^2=-2ab$.
$\displaystyle a^2 - b^2 = 1$ ...............(1)
$\displaystyle 2ab = -1$ ..................(2)
from (2), $\displaystyle b = - \frac 1{2a}$. plug that into (1) and solve for $\displaystyle a$. note that you must take the real answer. then plug that into (2) to solve for $\displaystyle b$