89. The value of the complex number
(1 + i)150 + (1 − i)150 is ______.
Find the Value of the Complex Number (1+i)150 + (1−i)150
Correct Answer
Answer: 0
Step 1: Express Each Complex Number in Polar Form
To evaluate the given expression, we first convert each complex number into its polar (trigonometric) form. For the complex number 1 + i, the modulus is calculated as
|1+i| = √(1² + 1²) = √2.
The argument of 1 + i is 45°, or π/4 radians, because both the real and imaginary parts are positive and equal. Therefore, its polar form is
1+i = √2 [cos(π/4) + i sin(π/4)].
Similarly, the modulus of 1 − i is also √2. However, its argument is −π/4 since the imaginary part is negative. Thus,
1−i = √2 [cos(−π/4) + i sin(−π/4)].
Step 2: Apply De Moivre’s Theorem
According to De Moivre’s Theorem, if a complex number is written as r(cos θ + i sin θ), then its nth power is given by
rn[cos(nθ) + i sin(nθ)].
Applying this theorem to the first complex number gives
(1+i)150 = (√2)150[cos(150π/4) + i sin(150π/4)].
Since (√2)150 = 275, the expression becomes
(1+i)150 = 275[cos(75π/2) + i sin(75π/2)].
Now simplify the angle.
75π/2 = 36π + 3π/2.
Because trigonometric functions are periodic with period 2π,
cos(75π/2) = cos(3π/2) = 0
and
sin(75π/2) = sin(3π/2) = −1.
Therefore,
(1+i)150 = −i·275.
Step 3: Evaluate the Second Term
Using the same theorem for the second complex number, we obtain
(1−i)150 = (√2)150[cos(−75π/2) + i sin(−75π/2)].
Using the identities cos(−θ) = cos θ and sin(−θ) = −sin θ, we find
cos(−75π/2) = 0
and
sin(−75π/2) = 1.
Hence,
(1−i)150 = i·275.
Step 4: Add the Two Terms
Finally, substitute the two evaluated expressions into the given equation.
(1+i)150 + (1−i)150
= −i·275 + i·275
= 0.
The imaginary terms are equal in magnitude but opposite in sign, so they cancel each other completely.
Why This Method Works
The two complex numbers, 1 + i and 1 − i, are complex conjugates of each other. Complex conjugates have the same modulus but opposite arguments. When both numbers are raised to the same power, their imaginary parts remain opposite, while their real parts remain identical. In this particular problem, the real parts become zero and the imaginary parts exactly cancel each other, leaving the final result equal to zero. This property is frequently exploited in competitive examinations to simplify otherwise lengthy calculations.
Important Concepts Used
Polar Form of a Complex Number
Every complex number can be represented as r(cos θ + i sin θ), where r is the modulus and θ is the argument. This representation makes it much easier to compute higher powers.
De Moivre’s Theorem
De Moivre’s Theorem is one of the most powerful tools in complex number theory. It converts repeated multiplication of complex numbers into a simple multiplication of angles, making questions involving large exponents straightforward.
Complex Conjugates
Two complex numbers are called conjugates if they have the same real part but opposite imaginary parts. Their powers often produce symmetric results, allowing many expressions to simplify dramatically.
Final Answer
After converting the complex numbers into polar form and applying De Moivre’s Theorem, we obtain
(1+i)150 + (1−i)150 = 0.
Final Answer: 0
