maths help: Complex Equations:

Start - Run - calc

Then i typed in
3 - (4 * x^2 - 1)

It may not be right, but make sure that when in windows calc you go to view and set to scientific
 
Not at all, I reckon the C i got in my Maths GCSE's was pure fluke, and I couldn't wait to get out of school and went straight to college and messed around for 2 years on a BTEC, But I did do on my HND a lot of business calculations mainly using Excel so I got to know my way around calc, very helpful bit of kit, even better my keyboard has a hotkey to it which I now use at work saves me loads a time.
 
(a^2-b^2)+a*b*j - so 3 = a^2-b^2 and -4 = a*b

EDIT: Think of it as factorising a quadratic - the roots will be of the form a+bj, so you just need to expand (a+bj)(a+bj)
 
One of the square roots is (2 - i).

If you were to do (2-i)(2-i) then you would get 4 + i^2 - 2i - 2i.
As i^2 = -1 you then get 4 - 1 - 4i = 3 - 4i.

Conversely the other root is (-2 + i).

(-2 + i)(-2 + i) = 4 + i^2 -2i -2i = 3 - 4i
 
Last edited:
Vixen said:
(a^2-b^2)+a*b*j - so 3 = a^2-b^2 and -4 = a*b

EDIT: Think of it as factorising a quadratic - the roots will be of the form a+bj, so you just need to expand (a+bj)(a+bj)
Click to expand...

thats what i thought, but why do we do that. thats what confuses me, we are sqrting yet somehow i have to factorise.

i mean yeah i could do that, and get my marks, but itll eat me till i know WHY thats what we do.
 
crooked i said:
thats what i thought, but why do we do that. thats what confuses me, we are sqrting yet somehow i have to factorise.

i mean yeah i could do that, and get my marks, but itll eat me till i know WHY thats what we do.
Click to expand...

It's because you're trying to find the sq root i.e. the number that is equal to your value (3-4j) when multiplied by itself:

(number)*(number) = 3 - 4j

To solve it you use the most generic thing you can - hence a + bj

Hope that makes it slightly clearer.
 
I got the square root as (5^0.5)*(cos(theta)+i.sin(theta))
with theta being -26.565 degrees.
EDIT: did you want the square root or roots of the equation?
EDIT2: actually it doesn't matter. 2-i is the same as ^
 
Last edited:
crooked i said:
thats what i thought, but why do we do that. thats what confuses me, we are sqrting yet somehow i have to factorise.

i mean yeah i could do that, and get my marks, but itll eat me till i know WHY thats what we do.
Click to expand...

To find the square root of a real number (n), you need to find a number (a), such that a*a = n. An imaginary number of the form a+bi will always have both a real (c) and imaginary (d) part to the root.
It is easier to think of the multiplication in the same way as factorising, so:
a+bi = (c+di)^2, ie a+bi = (c+di)(c+di) - expanding this bracket then gives what I posted before.

You just have to remember that there will always be two parts to the root and it will make more sense. Try it with some real numbers, for example a = 2, b = 3i and z = 2 + 3i. Square each of these, the only one that has two parts is the one that started off with two parts.
 
I'll go ahead with another example for you.

If you wanted to find the square roots of 15 + 8i this is what you would do.

If a + bi is a square root of 15 + 8i then:
(a + bi)² = 15 + 8i
therefore:
a² + 2abi - b² = 15 + 8i
(a² - b²) + 2abi = 15 + 8i

If we separate the imaginary and real parts we get for the real part
(a² - b²) = 15 & 2ab = 8
therefore a = 4/b

Substituting this value of a back into (a² - b²) = 15 we get:
(16/b²) - b² = 15
16 - b^4 = 15b²
b^4 + 15b² - 16 = 0
(b² + 16)(b² - 1) = 0

so b² must be either -16 or +1 (so that one of the backets = 0 so the sum can equal zero. However b² cannot be -16 as it wouldn't be real.

Therefore b² = 1 and therefore b = ±1

Thus going back to a = 4/b we get a = ±4

So the square roots of 15 + 8i are:

4 + i and -4 - i
 
You must log in or register to reply here.
Back
Top Bottom