Question

If z is a unimodular complex number such that Re(z - 1) + Re($z^2$) = $\int_{0}^{\pi/2}$ sin x ln | sin x - cos x | dx, then z can satisfy

• A. z + $\bar{z}$ = -2
• B. z + $\bar{z}$ = 1
• C. arg(z) = $\frac{\pi}{3}$
• D. arg z = $\pi$

Answer 1

Answer: (A), (B), (C), (D)

(1), (2), (3), (4)

Answer 2

The problem requires us to find which conditions a unimodular complex number $z$ can satisfy, given an equation involving its real part and an integral.

1. Evaluate the integral: Let the given integral be $I$. $I = \int_{0}^{\pi/2} \sin x \ln |\sin x - \cos x| dx$

Using the property $\int_a^b f(x) dx = \int_a^b f(a+b-x) dx$: $I = \int_{0}^{\pi/2} \sin(\pi/2 - x) \ln |\sin(\pi/2 - x) - \cos(\pi/2 - x)| dx$ $I = \int_{0}^{\pi/2} \cos x \ln |\cos x - \sin x| dx$

Since $|A| = |-A|$, we have $|\cos x - \sin x| = |\sin x - \cos x|$. So, $I = \int_{0}^{\pi/2} \cos x \ln |\sin x - \cos x| dx$.

Adding the two expressions for $I$: $2I = \int_{0}^{\pi/2} (\sin x + \cos x) \ln |\sin x - \cos x| dx$. Let $u = \sin x - \cos x$. Then $du = (\cos x + \sin x) dx$. When $x=0$, $u = \sin 0 - \cos 0 = 0 - 1 = -1$. When $x=\pi/2$, $u = \sin(\pi/2) - \cos(\pi/2) = 1 - 0 = 1$.

Substituting these into the integral: $2I = \int_{-1}^{1} \ln |u| du$. Since $\ln|u|$ is an even function, $\int_{-1}^{1} \ln |u| du = 2 \int_{0}^{1} \ln u du$. We know that $\int \ln u du = u \ln u - u$. So, $2I = 2 [u \ln u - u]_0^1 = 2 [(1 \ln 1 - 1) - \lim_{u \to 0^+} (u \ln u - u)]$. As $\ln 1 = 0$ and $\lim_{u \to 0^+} u \ln u = 0$: $2I = 2 [(0 - 1) - (0 - 0)] = 2(-1) = -2$. Therefore, $I = -1$.

2. Simplify the Left Hand Side (LHS): Given that $z$ is a unimodular complex number, its modulus is 1, i.e., $|z|=1$. Let $z = x + iy$, where $x, y \in \mathbb{R}$. From $|z|=1$, we have $x^2 + y^2 = 1$.

Re(z - 1) = Re((x - 1) + iy) = $x - 1$. $z^2 = (x + iy)^2 = x^2 - y^2 + 2ixy$. Re($z^2$) = $x^2 - y^2$. Substitute $y^2 = 1 - x^2$ (from $x^2+y^2=1$): Re($z^2$) = $x^2 - (1 - x^2) = x^2 - 1 + x^2 = 2x^2 - 1$.

Now, substitute these back into the LHS: LHS = Re(z - 1) + Re($z^2$) = $(x - 1) + (2x^2 - 1) = 2x^2 + x - 2$.

3. Equate LHS and RHS: We have LHS = $2x^2 + x - 2$ and RHS = $-1$. $2x^2 + x - 2 = -1$ $2x^2 + x - 1 = 0$. This is a quadratic equation in $x$. We can factor it: $(2x - 1)(x + 1) = 0$. This gives two possible values for $x = \text{Re}(z)$: $x = 1/2$ or $x = -1$.

4. Check the options based on possible values of $x$: Recall that for a complex number $z = x+iy$ with $|z|=1$, $x = \cos(\arg(z))$. Also, $z + \bar{z} = (x+iy) + (x-iy) = 2x$.

Case 1: $x = 1/2$ If $x = 1/2$, then $z + \bar{z} = 2(1/2) = 1$. This matches option (2). To find $y$, use $y^2 = 1 - x^2$: $y^2 = 1 - (1/2)^2 = 1 - 1/4 = 3/4$. So, $y = \pm \sqrt{3}/2$. The possible values for $z$ are $z = 1/2 + i\sqrt{3}/2$ or $z = 1/2 - i\sqrt{3}/2$. If $z = 1/2 + i\sqrt{3}/2$, then $z = e^{i\pi/3}$. So, $\arg(z) = \pi/3$. This matches option (3). If $z = 1/2 - i\sqrt{3}/2$, then $z = e^{-i\pi/3}$. So, $\arg(z) = -\pi/3$ (or $5\pi/3$).

Case 2: $x = -1$ If $x = -1$, then $z + \bar{z} = 2(-1) = -2$. This matches option (1). To find $y$, use $y^2 = 1 - x^2$: $y^2 = 1 - (-1)^2 = 1 - 1 = 0$. So, $y = 0$. The only possible value for $z$ is $z = -1 + i(0) = -1$. If $z = -1$, then $z = e^{i\pi}$. So, $\arg(z) = \pi$. This matches option (4).

Since the question asks "z can satisfy", it means any condition that holds for at least one valid $z$ is a correct option. We have found that:

• $z+\bar{z}=-2$ is satisfied by $z=-1$.
• $z+\bar{z}=1$ is satisfied by $z=1/2+i\sqrt{3}/2$ (or $z=1/2-i\sqrt{3}/2$).
• $\arg(z)=\pi/3$ is satisfied by $z=1/2+i\sqrt{3}/2$.
• $\arg(z)=\pi$ is satisfied by $z=-1$.

Therefore, all the given options are conditions that $z$ can satisfy.