Math question 8?

Question

Hi,
Could you solve the equation z^(2) = sqrt(3) - 3*i. Please answer in the form z = x + i*y.

* = multiply with
sqrt = square root

Answer 1

There will be 2 square roots.
Let's write √3 -3i in polar form:
We get r = √12 = 2√3
and arctan θ = -3/√3 = -√3,
θ = -π/3, because x is positive and y is negative.
When we take the square roots,
we have r = 12^(1/4)
and we must use θ/2 and θ/2+π as the arguments,
so θ = -π/6 and 5π/6.
Now we must convert both possibilities to rectangular form.
For the first case we get
x = 12^(1/4)*√3/2 and y = -12^(1/4)*/2.
The second case gives the negatives of x and y.
So z = ±12^(1/4)(√3/2 - i* /2)
Sorry for my earlier gaffes!

Answer 2

Okay. I think...but don't hold me to it....
that it will be z= 3^(1/4) - ( 3^(1/2) * i ^ (1/4) )
Hope that makes sense.
You're basically square-rooting the right hand side. So you raise each bit to the power of 1/2.

Answer 3

(x+ i y)^2 = sqrt(3) - 3*i

x^2 - y^2 + 2ixy =sqrt(3) - 3*i.

x^2-y^2 = sqrt(3)

xy = -3/2

x= 3/2 (fourth root 27/4)

y= fourth root 27/4

Answer 4

sqrt(3)-3i=z^2=x^2-y^2+2xyi so sqrt(3)=x^2-y^2
-3/2=xy
i leave the rest for you...solve this system...don't be too lazy

Answer 5

Is that your math homework you forgot to do little boy? and oh yeah how was chuckie cheese last night? Dont use the computer for your homework, how do you know people arent giving you the wrong answers?

Answer 6

isn't home work so you can learn and pass the class. not have others do the work for you.

Answer 7

It's summer why the hell are you doing math...

Answer 8

stop getting me to do your homework

Answer 9

goddness boyyyyyyy

Answer 10

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