Test: Complex Number- 2 - Airforce X Y / Indian Navy SSR MCQ

| 1 -i|^x = 2^x

we know,
|a + ib| = √(a² + b²) use this concept here,
so, |1 - i| = √{1² + (-1)²}
= √(1 + 1) = √2

now,
| 1 - i|^x = 2^x

(√2)^x = 2^x

(2½)^x = 2^x

2^(x/2) = 2^x

x/2 = x

x/2 - x = 0

x/2 = 0

x = 0

Let z = e-iθ
= cos(θ)−isin(θ)
arg(z)=−θ
arg(z)<0.
Now −z = −cos(θ)+isin(θ)
arg(−z) = π−θ
Hence arg(−z)−arg(z)
= π−θ−(−θ)
= π
 

ω is imaginary cube root of unity 

so 1 + ω + ω² = 1

ω² = -(1 +ω)

so our expression becomes (-ω² - ω²)⁷

= -128ω¹⁴

given ω³ = 1

so expresssion becomes -128ω²

Ar(z)= tan(1/√3) 
= π/6
since it lies in second quadrant
so it will be = π -π/6 
= 5π/6
 

A complex number z such that z ⁴=1. There are 4 fourth roots of unity and they are 1, i,−1 and−i.

 z³ – 2z² + 4z – 8 = 0
⇒ (z − 2)(z² + 4)=0
⇒ z − 2 = 0    : z = 2
⇒ z² + 4 = 0  : z = 2i, z = −2i
∴ z = 2, z = 2i, z = −2i
∴ ∣z∣ = 2

i = √-1
i = i
i² = (√-1)² = - 1
i³ = i² * i = - 1 * i  = - i
i⁴ = (i²)² = (-1)² = 1
i + i² + i³  + i⁴
= i - 1 - i + 1
= i - i + 1 - 1
= 0 + 0
= 0
i + i² + i³  + i⁴ = 0
⇒ i1 + i2 + i3 + i4 + ……… + i1000
⇒ 0 + 0 +........0(250 times)
⇒ 0

The sum of the three cube roots of unity is zero i.e., 

1+ω+ω²=0.

We know that, the sum of the three cube roots of unity 

The numbers are same on both sides but with a negative sign. So they will be same if (-1)^n is positive.

So values of n are even numbers.

smallest positive even number is 2.

Thus, value of n is 2.