Defining the Complex Conjugate
A complex number is typically expressed in the form $z = a + bi$, where $a$ and $b$ are real numbers and $i = \sqrt{-1}$.
The conjugate of a complex number $z$, denoted as $\bar{z}$, is obtained by changing the sign of the imaginary part. Therefore, if $z = a + bi$, then $\bar{z} = a - bi$.
Geometrically, the conjugate represents a reflection of the point $(a, b)$ across the real axis in the complex plane.
Solving the Problem
Step 1: Simplify the Complex Number
Before finding the square root, we must simplify the expression $\frac{2-36i}{2+3i}$. To do this, multiply the numerator and denominator by the conjugate of the denominator $(2-3i)$:
$$\frac{2-36i}{2+3i} \cdot \frac{2-3i}{2-3i} = \frac{4 - 6i - 72i + 108i^2}{2^2 + 3^2}$$
Since $i^2 = -1$:
$$\frac{4 - 78i - 108}{4 + 9} = \frac{-104 - 78i}{13} = -8 - 6i$$
Step 2: Finding the Square Root
Let $\sqrt{-8-6i} = x + iy$. Squaring both sides gives:
$$-8 - 6i = (x + iy)^2 = (x^2 - y^2) + 2xyi$$
Equating real and imaginary parts:
- $x^2 - y^2 = -8$
- $2xy = -6 \implies xy = -3$
Using the modulus property $|z|^2 = x^2 + y^2 = \sqrt{(-8)^2 + (-6)^2} = \sqrt{64 + 36} = 10$: 3) $x^2 + y^2 = 10$
Adding (1) and (3): $2x^2 = 2 \implies x^2 = 1 \implies x = \pm 1$
Subtracting (1) from (3): $2y^2 = 18 \implies y^2 = 9 \implies y = \pm 3$
Since $xy = -3$ (negative), $x$ and $y$ must have opposite signs. Thus, the square roots are: $$(1 - 3i) \text{ and } (-1 + 3i)$$
Summary
- The conjugate of $a+bi$ is $a-bi$.
- Always simplify complex fractions by multiplying by the conjugate of the denominator.
- To find square roots, equate the real and imaginary parts and solve the system of equations.