The correct answer is option (A): a$z$ + a$\bar{z}$ = 0
Let $a = \alpha + i\beta$, and $z = x + iy$.
Now, consider the given condition:
$\bar{a}z + a\bar{z} = 0$
Substituting the values:
$\Rightarrow (\alpha - i\beta)(x + iy) + (\alpha + i\beta)(x - iy) = 0$
This simplifies to a condition that represents a line.
Slope of the line: $-\frac{\alpha x}{\beta}$
Reflection slope: $\frac{\alpha}{\beta}x$
The line is: $\alpha x - \beta y = 0$, and the reflection also passes through the origin.
Using complex number representation:
$\Rightarrow \left(\frac{a + \bar{a}}{2}\right)\left(\frac{z + \bar{z}}{2}\right) - \left(\frac{a - \bar{a}}{2i}\right)\left(\frac{z - \bar{z}}{2i}\right) = 0$
Simplifies to: $az + \bar{a}z = 0$
Solve for \( x \):
\( \log_{10}(x^2) = 2 \).
Let \( K \) be an algebraically closed field containing a finite field \( F \). Let \( L \) be the subfield of \( K \) consisting of elements of \( K \) that are algebraic over \( F \).
Consider the following statements:
S1: \( L \) is algebraically closed.
S2: \( L \) is infinite.
Then, which one of the following is correct?
Consider z1 and z2 are two complex numbers.
For example, z1 = 3+4i and z2 = 4+3i
Here a=3, b=4, c=4, d=3
∴z1+ z2 = (a+c)+(b+d)i
⇒z1 + z2 = (3+4)+(4+3)i
⇒z1 + z2 = 7+7i
Properties of addition of complex numbers
It is similar to the addition of complex numbers, such that, z1 - z2 = z1 + ( -z2)
For example: (5+3i) - (2+1i) = (5-2) + (-2-1i) = 3 - 3i
Considering the same value of z1 and z2 , the product of the complex numbers are
z1 * z2 = (ac-bd) + (ad+bc) i
For example: (5+6i) (2+3i) = (5×2) + (6×3)i = 10+18i
Properties of Multiplication of complex numbers
Note: The properties of multiplication of complex numbers are similar to the properties we discussed in addition to complex numbers.
Associative law: Considering three complex numbers, (z1 z2) z3 = z1 (z2 z3)
Read More: Complex Numbers and Quadratic Equations
If z1 / z2 of a complex number is asked, simplify it as z1 (1/z2 )
For example: z1 = 4+2i and z2 = 2 - i
z1 / z2 =(4+2i)×1/(2 - i) = (4+i2)(2/(2²+(-1)² ) + i (-1)/(2²+(-1)² ))
=(4+i2) ((2+i)/5) = 1/5 [8+4i + 2(-1)+1] = 1/5 [8-2+1+41] = 1/5 [7+4i]