# [SOLVED] A complex number

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• September 14th 2008, 01:02 PM
arbolis
[SOLVED] A complex number
Find all the complex numbers such that $z^2=1-i$.
My attempt : I wrote $z=(a+ib)$ and with a little bit of some algebra I reached that $ab=-\frac{1}{2}$ which mean that if $ab=-\frac{1}{2}$ then $z^2=1-i$ is satisfied. So the solution is all z such that $\forall$ a,b $\in \mathbb{R}$ / $ab=-\frac{1}{2}$ with $z=a+ib$. 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.
• September 14th 2008, 01:11 PM
thelostchild
You are correct that
$2ab=-1$

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

so

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

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

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

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

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
$(1-\frac{1}{2}i)^2=1-\frac{1}{4}-1i=\frac{3}{4}-1i\neq 1-i$

therefore it cannot be correct on its own :)
• September 14th 2008, 01:32 PM
arbolis
I didn't forget to do that, the problem is that I didn't reached anything beautiful. Like $b=+$ or $-\sqrt{a^2-1}$, or $b^4-b^2=\frac{1}{4}$, or $4b^2(b^2+1)=1$ 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.
• September 14th 2008, 01:39 PM
Plato
Why not just find the two square roots of $1-i$?
• September 14th 2008, 01:39 PM
Jhevon
Quote:

Originally Posted by arbolis
I didn't forget to do that, the problem is that I didn't reached anything beautiful. Like $b=+$ or $-\sqrt{a^2-1}$, or $b^4-b^2=\frac{1}{4}$, or $4b^2(b^2+1)=1$ 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 $a = \pm \sqrt{\frac {1 + \sqrt{2}}2}$ and $b = - \frac 1{2a}$ ........and of course, you plug in what a is (can you get to that? :D)

you could also try converting to polar form, it might be the same, or more work, but not much different in terms of difficulty
• September 14th 2008, 02:04 PM
arbolis
I've reached $b=\pm \sqrt{\frac{1}{\pm 2\sqrt 2 +2}}$ so you can imagine to what $a$ is equal. But $b$ is a real number, so I can simply it to $\pm \sqrt{\frac{1}{2\sqrt 2 +2}}$. (Whew) I feel I made an error somewhere.
I've started with $a^2-b^2=-2ab$.
• September 14th 2008, 02:17 PM
Plato
In polar/exponential form $1 - i = \sqrt 2 \left( {\cos \left( {\frac{{ - \pi }}{4}} \right) + i\sin \left( {\frac{{ - \pi }}{4}} \right)} \right) = \sqrt 2 e^{i\left( {\frac{{ - \pi }}{4}} \right)}$
The roots are $z_1 = \sqrt[4]{2}\left( {\cos \left( {\frac{{ - \pi }}
{8}} \right) + i\sin \left( {\frac{{ - \pi }}{8}} \right)} \right)\;\& \;z_2 = \sqrt[4]{2}\left( {\cos \left( {\frac{{7\pi }}{8}} \right) + i\sin \left( {\frac{{7\pi }}{8}} \right)} \right)$
• September 14th 2008, 02:21 PM
arbolis
Quote:

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

Ah that's nice, I think I will do that in my exam even if I should simplify $-\frac{\pi}{8}$ and $\frac{7\pi}{8}$. I think my teacher will give me as much credit as if I solve it without using polar form.
• September 14th 2008, 02:28 PM
Jhevon
Quote:

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

without doing polar form, you should start with

$a^2 - b^2 = 1$ ...............(1)
$2ab = -1$ ..................(2)

from (2), $b = - \frac 1{2a}$. plug that into (1) and solve for $a$. note that you must take the real answer. then plug that into (2) to solve for $b$