Problem 7 - Algebra 1 Course in Mathematics for the IIT-JEE and Other Engineering Exams

Question

Find the square roots of the following (i) \(5+12 i\) (ii) \(-15-8 i\)

Step-by-Step Solution

Answer

\((5+12i)\) square roots are \(3 + 2i\) and \(-3 - 2i\); \((-15-8i)\) square roots are \(1 - 4i\) and \(-1 + 4i\).

Step 1: Understanding the Complex Number

We need to find the square root of the complex number for both parts (i) and (ii). Recall that for a complex number in the form of \( a+bi \), its square root is also a complex number and can be expressed as \( x+yi \).

Step 2: Equational Set-up (i)

For (i), let \( z = x + yi \) be the square root of \( 5 + 12i \). Then, \((x + yi)^2 = 5 + 12i\). Expanding, we have \(x^2 - y^2 + 2xyi = 5 + 12i\).

Step 3: Separating Real and Imaginary Parts (i)

Equate the real and imaginary parts: \(x^2 - y^2 = 5\) and \(2xy = 12\). From \(2xy = 12\), we get \(xy = 6\).

Step 4: Solving the System of Equations (i)

Using \(xy = 6\) and \(x^2 - y^2 = 5\), solve the system: \(x^2 = (x^2 - y^2) + y^2 = 5 + y^2\). Substitute \(y = \frac{6}{x}\) into this equation to find the values of \(x\) and \(y\).

Step 5: Calculating Solutions for (i)

Substitute \(y = \frac{6}{x}\) into the equation \(x^2 - \left(\frac{6}{x}\right)^2 = 5\), resulting in \(x^4 - 5x^2 - 36 = 0\). Solve for \(x\) using the quadratic equation \(x^2 = 9\) or \(x^2 = -4\). Use this to solve for \(x\) and \(y\): \(x = \pm 3, \; y = \pm 2i\). Therefore, the square roots are \(3 + 2i\) and \(-3 - 2i\).

Step 6: Equational Set-up (ii)

For (ii), let \( z = x + yi \) be the square root of \(-15 - 8i\). Then, \((x + yi)^2 = -15 - 8i\). Expanding, we have \(x^2 - y^2 + 2xyi = -15 - 8i\).

Step 7: Separating Real and Imaginary Parts (ii)

Equate the real and imaginary parts: \(x^2 - y^2 = -15\) and \(2xy = -8\). From \(2xy = -8\), we get \(xy = -4\).

Step 8: Solving the System of Equations (ii)

Using \(xy = -4\) and \(x^2 - y^2 = -15\), solve the system: \(x^2 = (x^2 - y^2) + y^2 = -15 + y^2\). Substitute \(y = \frac{-4}{x}\) into this equation to find the values of \(x\) and \(y\).

Step 9: Calculating Solutions for (ii)

Substitute \(y = \frac{-4}{x}\) into the equation \(x^2 - \left(\frac{-4}{x}\right)^2 = -15\), resulting in \(x^4 + 15x^2 - 16 = 0\). Solve the quadratic equation for \(x^2\): Use the quadratic formula to find \(x\) and \(y\), leading to \(x = \pm 1, y = \mp 4\). Therefore, the square roots are \(1 - 4i\) and \(-1 + 4i\).

Key Concepts

Square Roots

When we think of square roots, we typically associate them with real numbers. However, finding the square root of a complex number requires a deeper understanding. A complex number is a number that can be expressed in the form \(a + bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit, satisfying \(i^2 = -1\). The square root of such a number is also complex and can be represented as \(x + yi\). This means we're looking for values \(x\) and \(y\) such that \((x + yi)^2\) returns our original complex number. To solve for these, we expand \((x + yi)^2\) as \(x^2 - y^2 + 2xyi\). Then, we match the real part to the real part of the original complex number, and the imaginary part to the imaginary part, which provides us with a system of equations to solve.

System of Equations

Solving for the square root of complex numbers often boils down to solving a system of equations. When splitting a complex equation into its real and imaginary components, we arrive at two separate equations. For example, one equation represents the real parts, while the other represents the imaginary parts.

• For the real part: \(x^2 - y^2 = a\)
• For the imaginary part: \(2xy = b\)

These are the systems of equations you'll solve. These equations are typically nonlinear and require substitution or other algebraic methods to solve. By isolating one variable, substituting it into the other equation, and solving for the unknowns \(x\) and \(y\), you find the square roots of the complex number.

Quadratic Formula

The quadratic formula is an essential tool when dealing with complex numbers, especially when equations become too complex for simple algebraic manipulation. The quadratic formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]You'll use this formula to solve equations of the form \(ax^2 + bx + c = 0\). In the process of finding the complex square roots, you might end up with a polynomial equation, such as a fourth-degree polynomial. By rewriting it in terms of \(x^2\), you can apply the quadratic formula to find its solutions. This process will give you the values needed to determine your \(x\) and \(y\) values and hence find the required square root.

Real and Imaginary Parts

The real and imaginary parts of a complex number are integral to its identity. Whenever you deal with complex numbers, understanding how to separate and handle these parts is crucial. A complex number can always be split into:

• Real Part: This consists of numbers without the imaginary unit \(i\), represented as \(a\) in \(a + bi\).
• Imaginary Part: This consists of terms that include the imaginary unit \(i\), represented as \(b\) in \(a + bi\).

By separating a complex equation into its real and imaginary parts, you ensure that you maintain the balance and correlation between the two components. This technique is not only useful for solving square roots of complex numbers but is vital for any operation involving complex arithmetic. It allows a clearer representation that can be individually tackled and solved.