20 Important Examples: Conjugates of Complex Numbers (12th Grade)
Question 1: Write the conjugate of the complex number $z = 3 + 4i$.
Solution: The conjugate of a complex number $z = a + ib$ is formed by changing the sign of the imaginary part, giving $\bar{z} = a - ib$.
Therefore, $\bar{z} = 3 - 4i$.
Question 2: Write the conjugate of $z = -5 - 7i$.
Solution: By changing the sign of the imaginary component ($-7i$ becomes $+7i$), we get:
$\bar{z} = -5 + 7i$.
Question 3: Write the conjugate of $z = 8i$.
Solution: This is a purely imaginary number. It can be written as $z = 0 + 8i$. Reversing the sign of the imaginary part yields:
$\bar{z} = -8i$.
Question 4: Write the conjugate of $z = -12$.
Solution: This is a purely real number, which can be written as $z = -12 + 0i$. The conjugate only affects the imaginary part. Thus:
$\bar{z} = -12$.
Question 5: Write the conjugate of $z = 4i - 3$.
Solution: First, rewrite the complex number in standard form $a + ib$ to avoid confusion: $z = -3 + 4i$.
Now, change the sign of the imaginary part:
$\bar{z} = -3 - 4i$.
Question 6: Write the conjugate of $z = \sqrt{2} - i\sqrt{3}$.
Solution: Both the real and imaginary parts contain irrational numbers, but the rule remains the same. Flip the sign of the imaginary part:
$\bar{z} = \sqrt{2} + i\sqrt{3}$.
Question 7: Write the conjugate of $z = \frac{1}{2} + i\frac{\sqrt{3}}{2}$.
Solution: Changing the sign of the imaginary coefficient gives:
$\bar{z} = \frac{1}{2} - i\frac{\sqrt{3}}{2}$.
Question 8: Write the conjugate of $z = \cos \theta + i \sin \theta$.
Solution: For a complex number in trigonometric form, we simply negate the imaginary component:
$\bar{z} = \cos \theta - i \sin \theta$.
Question 9: Write the conjugate of $z = e^{i\pi/4}$.
Solution: In Euler's formula, the conjugate of $e^{i\theta}$ is $e^{-i\theta}$. Therefore:
$\bar{z} = e^{-i\pi/4}$.
Question 10: Write the conjugate of $z = \sqrt{-25}$.
Solution: First, simplify the radical into an imaginary number: $z = 5i$.
The conjugate of $5i$ is:
$\bar{z} = -5i$.
Question 11: Write the conjugate of $z = i^3$.
Solution: Recall that $i^2 = -1$, so $i^3 = i^2 \cdot i = -i$.
The conjugate of $-i$ is:
$\bar{z} = i$.
Question 12: Write the conjugate of $z = (2+3i) + (4-i)$.
Solution: First, simplify by adding the real parts and imaginary parts respectively: $z = (2+4) + i(3-1) = 6 + 2i$.
Now, find the conjugate:
$\bar{z} = 6 - 2i$.
Question 13: Write the conjugate of $z = -i(2 - 5i)$.
Solution: Distribute the $-i$:
$z = -2i + 5i^2$. Since $i^2 = -1$, we get $z = -2i - 5$, or in standard form, $z = -5 - 2i$.
The conjugate is:
$\bar{z} = -5 + 2i$.
Question 14: Write the conjugate of $z = (1+2i)(3-i)$.
Solution: First, expand the expression using FOIL:
$z = 3 - i + 6i - 2i^2$
$z = 3 + 5i - 2(-1)$
$z = 5 + 5i$.
The conjugate is:
$\bar{z} = 5 - 5i$.
Question 15: Write the conjugate of $z = (1-i)^2$.
Solution: Expand the binomial squared:
$z = 1^2 - 2(1)(i) + i^2$
$z = 1 - 2i - 1$
$z = -2i$.
The conjugate is:
$\bar{z} = 2i$.
Question 16: Write the conjugate of $z = \frac{1}{1+i}$.
Solution: First, express $z$ in standard form by multiplying the numerator and denominator by the conjugate of the denominator ($1-i$):
$z = \frac{1(1-i)}{(1+i)(1-i)} = \frac{1-i}{1^2 + 1^2} = \frac{1-i}{2} = \frac{1}{2} - \frac{1}{2}i$.
Now, take the conjugate of the result:
$\bar{z} = \frac{1}{2} + \frac{1}{2}i$.
Question 17: Write the conjugate of $z = \frac{3+2i}{2-5i}$.
Solution: Multiply the numerator and denominator by the conjugate of the denominator, $2+5i$:
$z = \frac{(3+2i)(2+5i)}{(2-5i)(2+5i)}$
$z = \frac{6 + 15i + 4i + 10i^2}{2^2 + 5^2}$
$z = \frac{6 + 19i - 10}{4 + 25} = \frac{-4 + 19i}{29} = -\frac{4}{29} + \frac{19}{29}i$.
The conjugate is:
$\bar{z} = -\frac{4}{29} - \frac{19}{29}i$.
Question 18: Write the conjugate of $z = 1 + i + i^2 + i^3$.
Solution: Simplify the powers of $i$ first:
$i^2 = -1$ and $i^3 = -i$.
$z = 1 + i - 1 - i = 0$.
The conjugate of $0$ is simply $0$.
$\bar{z} = 0$.
Question 19: Write the conjugate of $z = \frac{(1+i)^2}{3-i}$.
Solution: Simplify the numerator first: $(1+i)^2 = 1 + 2i + i^2 = 2i$.
Now, rationalize $z = \frac{2i}{3-i}$ by multiplying by $\frac{3+i}{3+i}$:
$z = \frac{2i(3+i)}{(3)^2 + (1)^2} = \frac{6i + 2i^2}{10} = \frac{-2 + 6i}{10} = -\frac{1}{5} + \frac{3}{5}i$.
The conjugate is:
$\bar{z} = -\frac{1}{5} - \frac{3}{5}i$.
Question 20: Write the conjugate of $z = x + iy$ (where $x, y \in \mathbb{R}$).
Solution: This represents the general algebraic form of any complex number. The conjugate is defined by replacing $+i$ with $-i$. Therefore:
$\bar{z} = x - iy$.