I know that complex numbers are written as a real number added to an imaginary number (i.e. -3+4i) but how do you square them?
If you can’t square complex numbers, then please explain why you can’t, and then explain how would this equation work out:
x=y^2+c, where y=z^2+c, z=0, and c is the complex number 0.5-0.75i
ddunx46135, does the equation
(a+b)^2=a^2+b^2+2ab only apply to complex numbers, or all numbers?
Square complex numbers by using (a + b)^2 = a^2 + 2ab + b^2 and then simplyfying by using i^2 = -1.
For example, (3 + 4i)^2 = 3^2 + 2(3)(4i) + (4i)^2
= 9 + 24i + 16(i^2)
= 9 + 24i -16
= -7 + 24i.
you square it by multiplying it by itself. this is accomplished by using the FOIL distribution rules, with the understanding that i^2=-1
References :
Square a complex number the same way you square any other binomial, use the distributive property:
(-3+4i)^2 = (-3+4i) (-3+4i)
= 9 – 24i – 16
= -7 – 24i
as for the second part just substitute 0.5-0.75i for c, and zero for z. then simplify.
Try it…hope this helps
References :
to square -3+4i u have to make it a product and foil:
(-3+4i) ( -3+4i) = 9 -12i -12i + 16i^2
however we know that i^2 = -1 so it becomes:
9 -24i-16 = -24i -7
References :
Same way you square a binomial:
(-3+4i)² = 9 – 24i + 16i²
= 9 – 24i – 16
= -7 -24i
—————
z=0
c = 0.5-0.75i
y=z²+c
= 0² + c
= c
x = y² + c
= c² + c
= c(c+1)
= (0.5-0.75i)(1.5-0.75i)
= 0.75 -0.375i – 1.125i + 0.5625i²
= 0.75 – 1.5i – 0.5625
= 0.1875 – 1.5i
References :
HI,
To square a complex number, write in out twice in 2 parentheses and then FOIL (multiply) them together. Remember that i² = -1.
(-3 + 4i)² =
(-3 + 4i)(-3 + 4i) =
9 – 12i – 12i + 16i² =
9 – 24i + 16(-1) =
9 – 24i – 16 =
-7 – 24i This is (-3 + 4i)²
If you have a graphing calculator, like a TI83, and you go under MODE and change from REAL to a + bi, your calculator will be able to work these out.
Enter (-3 + 4i)² into the calculator.
The "i" is above the decimal point key on the TI83.
Press ENTER.
Your answer appears!! -7 – 24i
That’s a nice way to double-check your work.
I hope that helps!!
References :
If the complex number is (a + i b) then the square is
a^2-b^2+i(2ab)
For example
=(-3+4i)^2
=(-3)^3+(4i)^2+i(2*(-3)*4)
=9-16-24*i
=-7-24*i
References :
You square complex number the same as squaring any binomial.
Thus, your (-3 + 4i)^2 = 9 -(2)(3)(4i) + (4i)^2 = 9 – 24i – 16
= – 7 – 24i
References :
Complex numbers can definitely be squared, and its quite simple. The biggest problem is just accepting i as a number, albeit imaginary, and treating is as such.
So, if we have an imaginary number a + bi, and we want to find (a + bi)^2, all we do is treat it like a binomial, expand, and simplify.
(a + bi)^2 is no more difficult to expand than (x + y)^2. The process is exactly the same.
Since (x + y)^2 = x^2 + 2xy + y^2, we have
(a + bi)^2 = a^2 + 2abi + (bi)^2.
But we’re not done here.
Remember, i is a number (albeit imaginary), so we will perform the same algebraic functions on it like we would any real number.
a^2 is real, so we can leave that alone.
2abi is an imaginary component, but cannot be simplified further.
(bi)^2 is the same as (b^2)*(i^2).
And since i = (-1)^(1/2),
i^2 = -1.
Substituing this for i^2 yields
(bi)^2 = -bi. So our final equation will be
a^2 + 2abi – b^2.
The inital example, (-3 + 4i)^2, would give us:
(-3)^2 + 2(-3)(4)i – (4)^2 = 9-24i-16 = -7 -24i
The problem with x, y, z, and c would be solved as follows:
x = y^2 + c
= (z^2 + c)^2 + c
= (0 + c)^2 + c
= c^2 + c.
Using the formula we just derived, we can now find c^2:
(0.5)^2 + 2(0.5)(-0.75)i – (0.75)^2, which equals
0.25+-0.75i-0.5625.
Adding c and combining our real and imaginary parts gives us:
0.25 + 0.5 – 0.5625 – 0.75i – 0.75i =
0.1875 – 1.5i.
References :
Square complex numbers by using (a + b)^2 = a^2 + 2ab + b^2 and then simplyfying by using i^2 = -1.
For example, (3 + 4i)^2 = 3^2 + 2(3)(4i) + (4i)^2
= 9 + 24i + 16(i^2)
= 9 + 24i -16
= -7 + 24i.
References :
Darrol, to your other question on the mandelbrot set (since I can’t answer a resolved question), yes, C is all possible values. That’s why you can draw a picture in two dimensions. Focus on a given point: x=0.3, y=0.2 (C = 0.3+0.2i), then determine if the set Z is bounded (it’s bounded), so then plot that point as black.
Squaring complex numbers is the same as squaring normal numbers. If you’re familiar with the F.O.I.L. method for squaring (first, outside, inside, last), (a+b)*(c+d) = first + outside + inside + last = (a*c) + (a*d) + (b*c) + (b*d). If (a+b)*(a+b) then you have (a*a) + (a*b) + (b*a) + (b*b) = a^2 + 2*a*b + b^2. (what ddun said)
So in your example: x=y^2+c, where y=z^2+c, z=0, and c is the complex number 0.5-0.75i
. . y=0^2+c=c
. . x=c^2+c
. . . = (0.5 – 0.75i) * (0.5 – 0.75i) + (0.5 – 0.75i)
. . . = (0.5*0.5 – 0.5*0.75i – 0.5*0.75i + 0.75i*0.75i) + (0.5 – 0.75i)
. . . = (0.25 – 0.375i – 0.375i + 0.5625*i^2) + (0.5 – 0.75i)
. . . = (0.25 – 0.375i – 0.375i + 0.5625*(-1)) + (0.5 – 0.75i)
. . . = (0.25 – 0.375i – 0.375i – 0.5625) + (0.5 – 0.75i)
. . . = (-0.3125-0.75i) + (0.5 – 0.75i)
. . . = 0.1875-1.5i (what D.W. and xaiym said)
So in the context of your Z series; Z={0,0.5-0.75i,0.1875-1.5i}.
I should probably point out that determing whether sets are "bounded" or not, is alot of work. I don’t think the mandelbrot set would have been invented without computers. It’s best to build the mandelbrot picture with a computer program.
References :
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