z = 1 + i√3 ← this is a complex number The mudulus of z is: m = √[(1)² + (√3)²] = √[1 + 3] = √4 = 2 The angle of z is β such as: tan(β) = (√3)/1 = √3 → β = π/3 When you mulitply 2 complex numbers together, you can get a new complex number: - the new modulus is the product of the 2 modulus - the new angle is the sum of the 2 angles Characteristics of z^(n) Its modulus is: m^(n) It's angle is: β * n …and we want to get: 64 ← new complex number → modulus = 64 and angle = 2π m^(n) = 64 → where; m = 2 2^(n) = 64 2^(n) = 2 * 2 * 2 * 2 * 2 * 2 2^(n) = 2^(6) n = 6 angle: (π/3) * n = 2π n = 2π/(π/3) n = 2π * (3/π) n = 6π/6 n = 6 So, we can see that: z^(6) = 64
Exponentiation in the complex plane is very simple if you use polar form. (r cis (theta))^n = r^n cis (n theta), where cis(theta) = cos theta + i sin theta. For the number have, r = sqrt(4) and theta=pi/3.
I'm getting it wrong and I can't see why (1+i√3)^n = |1+i√3|^n * e^(inπ/3) = 2^n * e^(inπ/3) = 64 2^n * e^(inπ/3) = 2^6 n*ln2 + inπ/3 = 6ln2 n = 6ln2 / (ln2 + iπ/3) which does not simplify to n = 6
z = 1 + i√3 ← this is a complex number
The mudulus of z is: m = √[(1)² + (√3)²] = √[1 + 3] = √4 = 2
The angle of z is β such as: tan(β) = (√3)/1 = √3 → β = π/3
When you mulitply 2 complex numbers together, you can get a new complex number:
- the new modulus is the product of the 2 modulus
- the new angle is the sum of the 2 angles
Characteristics of z^(n)
Its modulus is: m^(n)
It's angle is: β * n
…and we want to get: 64 ← new complex number → modulus = 64 and angle = 2π
m^(n) = 64 → where; m = 2
2^(n) = 64
2^(n) = 2 * 2 * 2 * 2 * 2 * 2
2^(n) = 2^(6)
n = 6
angle:
(π/3) * n = 2π
n = 2π/(π/3)
n = 2π * (3/π)
n = 6π/6
n = 6
So, we can see that: z^(6) = 64
Nice!
Absolute value method is great !
Very clean and accurate
|z| = |w| ==> Absolute values are ALWAYS equal 👌
Let x=e ^iπ/3,So (2x )^n=64,and x ^3=-1,so let n=3k,(-8) ^k=64,k=2,n=6
Exponentiation in the complex plane is very simple if you use polar form. (r cis (theta))^n = r^n cis (n theta), where cis(theta) = cos theta + i sin theta. For the number have, r = sqrt(4) and theta=pi/3.
Really like the 3d method. Very nice
Thanks so much 😊
The modulus of the lhs is 2 so n=6.
Would you say it's true that if there exists a real number such that the equation holds, it has to be n = 6 because 2^6 = 64?
I also got n=6.
Method 4:
(1+√3 i)^n=64
2^n(1/2+√3/2 i)n = 64
(-2w)^n =64
n=6
I'm getting it wrong and I can't see why
(1+i√3)^n = |1+i√3|^n * e^(inπ/3) = 2^n * e^(inπ/3) = 64
2^n * e^(inπ/3) = 2^6
n*ln2 + inπ/3 = 6ln2
n = 6ln2 / (ln2 + iπ/3) which does not simplify to n = 6
An Exponential Equation: (1 + i√3)^n = 64; n =?
Let: a = 1 + i√3; a² = (1 + i√3)² = - 2 + 2i√3 = - 2 + 2(a - 1) = 2a - 4
a³ = a(a²) = a(2a - 4) = 2a² - 4a = 2(2a - 4) - 4a = - 8, a⁶ = (a³)² = (- 8)² = 64
(1 + i√3)^n = a^n = 64 = a⁶; n = 6
Answer check:
n = 6: (1 + i√3)^6 = 64; Confirmed as shown
Final answer:
n = 6