What is the value of \(\sqrt{12+5 i}+\sqrt{12-5 i}\) where \(i=\sqrt{-1}\) ?
5√2
Understanding the Complex Number Problem
The question asks us to find the value of the expression \( \sqrt{12+5 i}+\sqrt{12-5 i} \), where \(i\) is the imaginary unit defined as \( \sqrt{-1} \). This problem involves complex numbers and their square roots.
The two complex numbers inside the square roots, \(12+5i\) and \(12-5i\), are complex conjugates. The sum of the square roots of complex conjugates usually results in a real number.
Step-by-Step Calculation
Let the given expression be equal to \(X\):
\( X = \sqrt{12+5 i}+\sqrt{12-5 i} \)
To eliminate the square roots, we can square both sides of the equation:
\( X^2 = \left( \sqrt{12+5 i}+\sqrt{12-5 i} \right)^2 \)
We use the algebraic identity \( (a+b)^2 = a^2 + b^2 + 2ab \), where \( a = \sqrt{12+5 i} \) and \( b = \sqrt{12-5 i} \).
\( X^2 = (\sqrt{12+5 i})^2 + (\sqrt{12-5 i})^2 + 2 \sqrt{12+5 i} \sqrt{12-5 i} \)
Simplifying the squared terms:
- \( (\sqrt{12+5 i})^2 = 12+5i \)
- \( (\sqrt{12-5 i})^2 = 12-5i \)
Now consider the product term \( 2 \sqrt{12+5 i} \sqrt{12-5 i} \). We can combine the terms under a single square root since they are being multiplied:
\( 2 \sqrt{(12+5 i)(12-5 i)} \)
The product inside the square root is of the form \( (a+bi)(a-bi) \), which simplifies to \( a^2 + b^2 \). Here, \(a=12\) and \(b=5\).
\( (12+5 i)(12-5 i) = 12^2 + 5^2 = 144 + 25 = 169 \)
So, the product term becomes:
\( 2 \sqrt{169} = 2 \times 13 = 26 \)
Now substitute these simplified terms back into the equation for \(X^2\):
\( X^2 = (12+5i) + (12-5i) + 26 \)
\( X^2 = 12 + 5i + 12 - 5i + 26 \)
The imaginary parts \(+5i\) and \(-5i\) cancel each other out:
\( X^2 = (12 + 12) + (5i - 5i) + 26 \)
\( X^2 = 24 + 0 + 26 \)
\( X^2 = 50 \)
To find \(X\), we take the square root of both sides:
\( X = \sqrt{50} \)
Simplify the square root of 50:
\( \sqrt{50} = \sqrt{25 \times 2} = \sqrt{25} \times \sqrt{2} = 5\sqrt{2} \)
Since \(X\) is the sum of the principal square roots of a complex number and its conjugate, \(X\) must be a non-negative real number. Therefore, we take the positive square root.
\( X = 5\sqrt{2} \)
Summary of Calculation
We found that squaring the expression \( \sqrt{12+5 i}+\sqrt{12-5 i} \) resulted in 50. Taking the square root of 50 gives \(5\sqrt{2}\).
| Step | Calculation | Result |
|---|---|---|
| 1 | Define \(X\) | \(X = \sqrt{12+5 i}+\sqrt{12-5 i}\) |
| 2 | Square \(X\) | \(X^2 = (\sqrt{12+5 i}+\sqrt{12-5 i})^2\) |
| 3 | Expand \(X^2\) | \(X^2 = (12+5i) + (12-5i) + 2\sqrt{(12+5i)(12-5i)}\) |
| 4 | Calculate product of conjugates | \((12+5i)(12-5i) = 12^2 + 5^2 = 169\) |
| 5 | Simplify product term | \(2\sqrt{169} = 2 \times 13 = 26\) |
| 6 | Substitute back into \(X^2\) | \(X^2 = (12+5i) + (12-5i) + 26\) |
| 7 | Simplify \(X^2\) | \(X^2 = 24 + 26 = 50\) |
| 8 | Find \(X\) | \(X = \sqrt{50}\) |
| 9 | Simplify \(\sqrt{50}\) | \(X = 5\sqrt{2}\) |
Revision Table: Complex Numbers & Square Roots
| Concept | Description | Formula/Property |
|---|---|---|
| Complex Number | A number of the form \(a+bi\), where \(a\) and \(b\) are real numbers, and \(i\) is the imaginary unit (\(\sqrt{-1}\)). | \(z = a+bi\) |
| Imaginary Unit | Defined as the square root of \(-1\). | \(i = \sqrt{-1}\), \(i^2 = -1\) |
| Complex Conjugate | For a complex number \(a+bi\), its conjugate is \(a-bi\). | Conjugate of \(z=a+bi\) is \(\bar{z} = a-bi\) |
| Product of Conjugates | The product of a complex number and its conjugate is a real number, equal to the sum of the squares of its real and imaginary parts. | \((a+bi)(a-bi) = a^2 + b^2\) |
| Square Root of a Real Number | For a positive real number \(x\), \(\sqrt{x}\) denotes the principal (non-negative) square root. | \( \sqrt{x} \ge 0 \) for \( x \ge 0 \) |
| Square Root of a Complex Number | A complex number \(z\) has two square roots. The principal square root is typically denoted by \(\sqrt{z}\). | \(\sqrt{z}\) is a complex number \(w\) such that \(w^2 = z\). |
Additional Information on Complex Number Square Roots
Finding the square root of a complex number \(a+bi\) involves solving for real numbers \(x\) and \(y\) such that \( (x+yi)^2 = a+bi \). This expands to \( x^2 - y^2 + 2xyi = a+bi \).
Equating the real and imaginary parts gives a system of equations:
- \( x^2 - y^2 = a \)
- \( 2xy = b \)
Also, using the magnitude: \( |(x+yi)^2| = |a+bi| \), which gives \( x^2 + y^2 = \sqrt{a^2+b^2} \).
Solving these equations simultaneously yields the values for \(x\) and \(y\). For the principal square root, we usually choose the root with a positive real part, or if the real part is zero, a positive imaginary part.
In this problem, calculating \(\sqrt{12+5i}\) and \(\sqrt{12-5i}\) individually is more complex than squaring the sum directly, as shown in the step-by-step solution. The direct method leverages the properties of complex conjugates and square roots.
The final value of the expression \( \sqrt{12+5 i}+\sqrt{12-5 i} \) is \( 5\sqrt{2} \).
Important Questions from Properties of Complex Numbers
If z z̅ = |z + z̅ |, where z = x + iy, i = \(\sqrt{-1}\), then the locus of z is a pair of:
If z is a complex number such that \(\frac{z-1}{z+1}\) is purely imaginary, then what is |z| equal to ?
What is the real part of (sin x + icos x) 3
What is z 1+ z 2+ z 3equal to?
Consider the following statements:
1. z 1 z 2 z 3 is purely imaginary.
2. z 1z 2 + z 2z 3 + z 3z 1 is purely real.
Which of the above statements is/are correct?
