# Complex Numbers - Find z & w

• Jul 17th 2011, 04:33 AM
Flamuri
Complex Numbers - Find Z & W
Assign Z & W [ Complex numbers ] if

$z + w = 1 - i$ ;

$argZ = \frac{\pi}{6}$ ;

$argW = \frac{5\pi }{3}$

Thanks,
Flamuri
• Jul 17th 2011, 04:49 AM
Prove It
Re: Complex Numbers - Find Z & W
Well, we can see that $\displaystyle z = r_1\left(\cos{\frac{\pi}{6}} + i\sin{\frac{\pi}{6}}\right) = r_1\left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right) = \frac{\sqrt{3}}{2}r_1 + \frac{1}{2}ir_1$ and $\displaystyle w = r_2\left(\cos{\frac{5\pi}{3}} + i\sin{\frac{5\pi}{3}}\right) = r_2\left(\frac{1}{2} - \frac{\sqrt{3}}{2}i\right) = \frac{1}{2}r_2 - \frac{\sqrt{3}}{2}ir_2$.

So $\displaystyle z + w = \frac{\sqrt{3}}{2}r_1 + \frac{1}{2}r_2 + \left(\frac{1}{2}r_1 - \frac{\sqrt{3}}{2}r_2\right)i$.

Now equate your real and imaginary parts to give you two equations to solve for $\displaystyle r_1$ and $\displaystyle r_2$ and you will have $\displaystyle z$ and $\displaystyle w$.
• Jul 17th 2011, 04:57 AM
Siron
Re: Complex Numbers - Find Z & W
modified.
• Jul 17th 2011, 05:39 AM
Flamuri
Re: Complex Numbers - Find Z & W
Wow, I've used a completely different way [for me a bit harder way] from this that you have used, but finally found a result:

If

$z = a + bi$
$w = x + yi$

So:

$z = - \frac{\sqrt{3}+3}{2} + \frac{\sqrt{3}-1}{2}i$

$w = \frac{\sqrt{3}+3}{2} + \frac{\sqrt{3}+1}{2}i$

Don't know how to prove if i'm right, maybe you can help me again,
Thanks.
• Jul 17th 2011, 05:43 AM
Plato
Re: Complex Numbers - Find Z & W
Quote:

Originally Posted by Flamuri
Assign Z & W [ Complex numbers ] if
$z + w = 1 - i$ ;

$argZ = \frac{\pi}{6}$ ; $argW = \frac{5\pi }{3}$

I would do it this way. Let $z=a+bi~\&~w=c+di$.

From the given \begin{align*} a+c&= 1 \\ b+d &=-1 \end{align*} and $\frac{b}{a} = \tan \left( {\frac{\pi }{6}} \right)\;\& \,\frac{d}{c} = \tan \left( {\frac{{5\pi }}{3}} \right)$
• Jul 17th 2011, 06:10 AM
Prove It
Re: Complex Numbers - Find Z & W
Quote:

Originally Posted by Flamuri
Wow, I've used a completely different way [for me a bit harder way] from this that you have used, but finally found a result:

If

$z = a + bi$
$w = x + yi$

So:

$z = - \frac{\sqrt{3}+3}{2} + \frac{\sqrt{3}-1}{2}i$

$w = \frac{\sqrt{3}+3}{2} + \frac{\sqrt{3}+1}{2}i$

Don't know how to prove if i'm right, maybe you can help me again,
Thanks.

I don't know what you've done here... You're told $\displaystyle z + w = 1 - i$ and we've found

$\displaystyle z + w = \frac{\sqrt{3}}{2}r_1 + \frac{1}{2}r_2 + \left(\frac{1}{2}r_1 - \frac{\sqrt{3}}{2}r_2\right)i$.

Surely that means

$\displaystyle \frac{\sqrt{3}}{2}r_1 + \frac{1}{2}r_2 = 1$ and $\displaystyle \frac{1}{2}r_1 - \frac{\sqrt{3}}{2}r_2 = -1$.

Solve these equations simultaneously for $\displaystyle r_1$ and $\displaystyle r_2$.