Author Topic: complex numbers question (Read 826 times) Tweet Share

olly_s15 · « **on:** November 11, 2009, 07:55:26 pm »

Solve for real values x and y:

$(x+iy)^2 = -18i$

TrueTears · « **Reply #1 on:** November 11, 2009, 07:57:57 pm »

Best way is to use De moivres theorem.

Let $z = x+yi = rCis(\theta)$

Now $-18i = 18Cis(\frac{-\pi}{2})$

So $r^2Cis(2\theta) = 18Cis(\frac{-\pi}{2})$

Can you do it from there?

olly_s15 · « **Reply #2 on:** November 11, 2009, 08:13:15 pm »

Oh, i'm teaching myself spesh to get ahead for next year so i haven't covered that yet. I'm using essentials and this is the first complex numbers exercise so yeah, i think it wants us to do it algebraically but i'm stuck

TrueTears · « **Reply #3 on:** November 11, 2009, 08:18:50 pm »

Ahh okay, well what you do is you expand $(x+yi)^2$ and then equate real and imaginary parts.

olly_s15 · « **Reply #4 on:** November 11, 2009, 08:46:32 pm »

i can't seem to get it, i end up with equating x^2=y^2 and 2xy=-18

and this fails

TrueTears · « **Reply #5 on:** November 11, 2009, 08:48:01 pm »

$x^2 + 2xyi + (yi)^2 = -18i$

$x^2 + 2xyi -y^2 = -18i$

Thus

$x^2-y^2 = 0 ...[1]$

$2xy = -18 ...[2]$

Solve the simultaneous equation.

olly_s15 · « **Reply #6 on:** November 11, 2009, 08:53:14 pm »

tried several times

i can't do it it's frustrating me

TrueTears · « **Reply #7 on:** November 11, 2009, 08:56:38 pm »

from [2]

$y = \frac{-9}{x}$

Sub in [1]

$x^2 - \frac{81}{x^2} = 0$

$x^4 - 81 = 0$

$(x^2)^2 - (3^2)^2 = 0$

$(x^2 + 3^2)(x^2 - 3^2) = 0$

$x^2 = 9$

$x = \pm 3$

Can you work out y?

Thus the reason I dislike this method, too much algebra, you'll love de moivres when you get up to it, so elegant ^.^

olly_s15 · « **Reply #8 on:** November 11, 2009, 09:01:19 pm »

O M G i made a stupid mistake
thanks haha it's not hard now that i see my mistake omgggg

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