Let S be the set of all (α, β), π

Q: Why do we reject the positive root for $\sin \alpha$?

Great question! We are constrained by the domain $\pi < \alpha < 2\pi$. In the unit circle, this corresponds to the third and fourth quadrants. In these quadrants, the sine function is strictly negative. Thus, even though the algebra gives $\sin \alpha = \pm \frac{1}{\sqrt{2}}$, we must discard the positive value.

Q: Is there a faster way to check if a number is purely imaginary without full rationalization?

Yes! A complex number $w$ is purely imaginary if and only if $w + \bar{w} = 0$. Similarly, $u$ is purely real if $u = \bar{u}$. While this can sometimes be faster, rationalization is often safer to avoid algebraic errors when the expressions are complex.

Q: Why does the summation term $i Z_{\alpha\beta} + \frac{1}{i \bar{Z}_{\alpha\beta}}$ look so intimidating?

It looks scary, but don't let the notation intimidate you. Notice that $\frac{1}{i} = -i$. So the expression is actually $i Z - \frac{i}{\bar{Z}}$. Once you calculate the specific values of $Z$ for your pairs of $(\alpha, \beta)$, the arithmetic becomes much simpler than it appears.

Question

Let $S$ be the set of all $(\alpha, \beta)$, $\pi < \alpha, \beta < 2\pi$, for which the complex number $\frac{1 - i\sin \alpha}{1 + 2i\sin \alpha}$ is purely imaginary and $\frac{1 + i\cos \beta}{1 - 2i\cos \beta}$ is purely real. Let $Z_{\alpha\beta} = \sin 2\alpha + i\cos 2\beta, (\alpha, \beta) \in S$. Then $\sum_{(\alpha, \beta) \in S} (i Z_{\alpha\beta} + \frac{1}{i \bar{Z}_{\alpha\beta}})$ is equal to:

Options:

(A) 3

(B) 3i

(C) 1

(D) 2 - i

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given two angles, $\alpha$ and $\beta$, both constrained within the interval $(\pi, 2\pi)$. Our objective is to evaluate a specific expression involving these angles by first determining their values through complex number conditions.

### Phase 1: The Purely Imaginary Condition

Let us focus on the expression $w = \frac{1 - i\sin \alpha}{1 + 2i\sin \alpha}$. For $w$ to be purely imaginary, its real part must be zero. We rationalize the denominator by multiplying by its conjugate, $1 - 2i\sin \alpha$:

$$w = \frac{(1 - i\sin \alpha)(1 - 2i\sin \alpha)}{(1 + 2i\sin \alpha)(1 - 2i\sin \alpha)} = \frac{1 - 2i\sin \alpha - i\sin \alpha + 2i^2\sin^2 \alpha}{1 + 4\sin^2 \alpha}$$

Using the identity $i^2 = -1$, the real part is given by:

$$	ext{Re}(w) = \frac{1 - 2\sin^2 \alpha}{1 + 4\sin^2 \alpha}$$

Setting $	ext{Re}(w) = 0$ implies $1 - 2\sin^2 \alpha = 0$, or $\sin^2 \alpha = \frac{1}{2}$. Given $\pi < \alpha < 2\pi$, we must have $\sin \alpha = -\frac{1}{\sqrt{2}}$, which yields $\alpha = \frac{5\pi}{4}$ and $\alpha = \frac{7\pi}{4}$.

### Phase 2: The Purely Real Condition

Next, we consider $u = \frac{1 + i\cos \beta}{1 - 2i\cos \beta}$. For $u$ to be purely real, its imaginary part must be zero. Rationalizing the expression:

$$u = \frac{(1 + i\cos \beta)(1 + 2i\cos \beta)}{(1 - 2i\cos \beta)(1 + 2i\cos \beta)} = \frac{1 + 2i\cos \beta + i\cos \beta + 2i^2\cos^2 \beta}{1 + 4\cos^2 \beta}$$

The imaginary part is:

$$	ext{Im}(u) = \frac{3\cos \beta}{1 + 4\cos^2 \beta}$$

Setting $	ext{Im}(u) = 0$ requires $\cos \beta = 0$. Within the domain $\pi < \beta < 2\pi$, the only solution is $\beta = \frac{3\pi}{2}$.

### Phase 3: The Final Summation

We define $Z_{\alpha\beta} = \sin 2\alpha + i\cos 2\beta$. We evaluate $Z$ for our pairs $(\alpha, \beta)$:

For $\alpha = \frac{5\pi}{4}$, $Z_1 = \sin(\frac{5\pi}{2}) + i\cos(3\pi) = 1 - i$.

For $\alpha = \frac{7\pi}{4}$, $Z_2 = \sin(\frac{7\pi}{2}) + i\cos(3\pi) = -1 - i$.

We evaluate $f(Z) = iZ + \frac{1}{i\bar{Z}} = iZ - \frac{i}{\bar{Z}}$. Substituting our values:

$$f(Z_1) = i(1 - i) - \frac{i}{1 + i} = (1 + i) - \frac{i(1 - i)}{2} = \frac{1 + i}{2}$$

$$f(Z_2) = i(-1 - i) - \frac{i}{-1 + i} = (1 - i) - \frac{i(-1 - i)}{2} = \frac{1 - i}{2}$$

Summing these results, we obtain:

$$\frac{1 + i}{2} + \frac{1 - i}{2} = \mathbf{1}$$

Select Answer:

Visualized Solution

Problem Orientation

Condition for Purely Imaginary

Rationalizing the First Expression

Setting Real Part to Zero

Solving for $α$

Condition for Purely Real

Rationalizing the Second Expression

Setting Imaginary Part to Zero

Solving for $β$

Defining Set $S$ and $Z_{α β}$

Calculating $Z_{1}$ and $Z_{2}$

Simplifying the Summation Term

Evaluating Term for $Z_{1}$

Evaluating Term for $Z_{2}$

Final Summation

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Similar Questions

Let $A = {θ \in (0, 2 π) : \frac{1 + 2 i s i n θ}{1 - i s i n θ} is purely imaginary}$ . Then the sum of the elements in $A$ is

Let $A = {θ \in (- π /2, π) : \frac{3 + 2 i s i n θ}{1 - 2 i s i n θ} is purely imaginary}$ . Then the sum of the elements in A is :

The sum of all possible values of $θ \in [- π, 2 π]$ , for which $\frac{1 + i c o s θ}{1 - 2 i c o s θ}$ is purely imaginary, is equal

Sum of squares of modulus of all the complex numbers $z$ satisfying $\overset{z}{ˉ} = i z^{2} + z^{2} - z$ is equal to \_\_\_\_\_.

Let $S = {z \in C : z^{2} + \overset{z}{ˉ} = 0}$ . Then $\sum_{z \in S} (Re (z) + Im (z))$ is equal to

Let $S = {z \in C : \overset{z}{ˉ} = i (z^{2} + Re (\overset{z}{ˉ}))}$ . Then $\sum_{z \in S} ∣ z ∣^{2}$ is equal to

Let $A = {θ \in [0, 2 π] : 1 + 10 Re (\frac{2 c o s θ + i s i n θ}{c o s θ - 3 i s i n θ}) = 0}$ . Then $\sum_{θ \in A} θ^{2}$ is equal to

If $z = \frac{1}{2} - 2 i$ , is such that $∣ z + 1∣ = α z + β (1 + i), i = - 1$ and $α, β \in R$ , then $α + β$ is equal to

Let $z \in C$ be such that $\frac{z ^{2} + 3 i}{z - 2 + i} = 2 + 3 i$ . Then the sum of all possible values of $z^{2}$ is

For any two complex numbers $z_{1}, z_{2}$ and any real number $a$ and $b$ , $∣ a z_{1} - b z_{2} ∣^{2} + ∣ b z_{1} + a z_{2} ∣^{2} = \dots$

Select Answer:

Visualized Solution

Problem Orientation

Condition for Purely Imaginary

Rationalizing the First Expression

Setting Real Part to Zero

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Condition for Purely Real

Rationalizing the Second Expression

Setting Imaginary Part to Zero

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β

Defining Set .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } S and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Zαβ​

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Z1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Z2​

Simplifying the Summation Term

Evaluating Term for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Z1​

Evaluating Term for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Z2​

Final Summation

The Sigma Insight: Algebraic Operations on Complex Numbers

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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