I need help converting the complex number z = (-1 + sqrt(1 + 4i))/2i to the form rei^(theta).
I found what the number in the square root is in the form rei^(theta) is which is sqrt(1 + 4i) = sqrt(17) e^i(tan-1(4) + pi*k) , but I am having trouble dealing with i in the denominator of z.
1+4i = √17 cisθ where tanθ=4
so, √(1+4i) = ∜17 cis(θ/2) = 2.0305 e^0.6629i = 1.6005+1.2496i
So, we have
-1+√(1+4i) = 0.6005+1.2496i
(-1+√(1+4i))/2i = 0.3002/i + 0.6248
= 0.6248-0.3002i
Now just convert that to polar form.
Thank you!
To convert the complex number z = (-1 + sqrt(1 + 4i))/2i to the form re^(iθ), we can follow these steps:
Step 1: Simplify the expression inside the square root:
- Start with 1 + 4i
- Convert it to polar form by finding the magnitude and argument.
- The magnitude (r) can be found using the formula r = sqrt(a^2 + b^2), where a and b are the real and imaginary parts respectively.
- In this case, a = 1 and b = 4, so r = sqrt(1^2 + 4^2) = sqrt(1 + 16) = sqrt(17).
- To find the argument (θ), we can use the formula tan(θ) = b/a. This gives us tan(θ) = 4/1, so θ = tan^(-1)(4).
- Now we have the complex number sqrt(1 + 4i) in the form re^(iθ), which is sqrt(17) e^(i(tan^(-1)(4))).
Step 2: Simplify the denominator of z:
- The denominator is 2i.
- We can multiply it by its conjugate, which is -2i, to get a real number in the denominator.
- Multiply the numerator and denominator of z by -2i. This gives us:
z = (-1 + sqrt(1 + 4i))/2i * (-2i)/(-2i)
= (2i - 2i*sqrt(1 + 4i)) / (-4)
= (-2i + 2i*sqrt(1 + 4i))/4
= (2i*sqrt(1 + 4i) - 2i)/4
= (i*sqrt(1 + 4i) - i)/2
Step 3: Combine the numerator and denominator of z:
- Now we can combine the numerator and denominator.
Numerator: i*sqrt(1 + 4i) - i
= i*sqrt(17) e^(i(tan^(-1)(4))) - i
Denominator: 2
So, z = (i*sqrt(1 + 4i) - i)/2.
Overall, z is equal to (i*sqrt(1 + 4i) - i)/2 in the desired form re^(iθ).