Write down 3 other possible trig forms for - 1 + isqrt2. How do you solve this?
To find the three other possible trigonometric forms for the complex number -1 + i√2, we need to first express the number in its standard form a + bi, where a and b represent the real and imaginary parts, respectively.
Given the number -1 + i√2, we can rewrite it as -1 + i√(2) (since the square root of 2 cannot be simplified further).
To find the standard form, we need to rationalize the denominator of i√(2). Multiplying the numerator and denominator by the conjugate of the denominator will help eliminate the square root.
Thus, we can rewrite the number as follows:
-1 + i√(2) * (i√(2) / i√(2))
= -1 + i * √(2) * i / √(2) * i
= -1 + i^2 * √(2) / i^2 * √(2)
= -1 - √(2) / -1 * √(2)
Now we have -1 - √(2) / -√(2). Simplifying further:
= -√(2) * (√(2) + 1) / 2
= -(√(2) * √(2) + √(2)) / 2
= -(2 + √(2)) / 2
= -1 - (√(2)/2)
Now, let's find the trigonometric forms of this complex number:
1) Polar Form:
To convert the number to polar form, we need to find the magnitude (r) and argument (θ).
The magnitude (r) can be calculated using the formula:
r = |a + bi| = √(a^2 + b^2)
For the given number:
r = √((-1)^2 + (√(2)/2)^2)
= √(1 + 2/4)
= √(1.5)
= √(3/2)
= √(3) / √(2)
= (√(3) * √(2)) / (√(2) * √(2))
= (√(6)) / 2
The argument (θ) can be calculated using the formula:
θ = arctan(b / a)
For the given number:
θ = arctan((√(2)/2) / -1)
= arctan(-√(2) / 2)
= - π/4
Therefore, in polar form, -1 + i√2 can be written as (√6 / 2) * (cos(-π/4) + i * sin(-π/4)).
2) Exponential Form:
To express the complex number in exponential or Euler's form, we can use the formula:
z = re^(iθ)
For the given number:
z = (√6 / 2) * e^(i * (- π/4))
= (√6 / 2) * cos(-π/4) + i * sin(-π/4)
3) Trigonometric Form:
The trigonometric form is typically written as r(cos(θ) + i * sin(θ)).
For the given number:
x = (√6 / 2) * cos(-π/4)
y = (√6 / 2) * sin(-π/4)
Thus, the three other possible trigonometric forms for -1 + i√2 are:
1) (√6 / 2) * (cos(-π/4) + i * sin(-π/4))
2) (√6 / 2) * (cos(7π/4) + i * sin(7π/4))
3) (√6 / 2) * (cos(15π/4) + i * sin(15π/4))
Note: In mathematics, there is an infinite number of possible angles for a given complex number in trigonometric form, as they repeat every 2π due to the periodic nature of the trigonometric functions.