Chapter 1: Problem 49
Assume for the moment that \(\sqrt{1+i}\) makes sense in the complex number system. How would you then demonstrate the validity of the equality $$ \sqrt{1+i}=\sqrt{\frac{1}{2}+\frac{1}{2} \sqrt{2}}+i \sqrt{-\frac{1}{2}+\frac{1}{2} \sqrt{2}} ? $$
Short Answer
Expert verified
The equality is valid, as shown by expressing \(\sqrt{1+i}\) as \(a + bi\) and solving for \(a\) and \(b\).
Step by step solution
01
Assume the Form of the Solution
Assume the solution in the form \( \sqrt{1+i} = a + bi \), where \(a\) and \(b\) are real numbers. Our goal is to determine the values of \(a\) and \(b\).
02
Square Both Sides
Square both sides of the equation \((a + bi)^2 = 1 + i\). This gives:\[a^2 - b^2 + 2abi = 1 + i\]
03
Equate Real and Imaginary Parts
Equate the real and imaginary parts of the equation:- Real part: \(a^2 - b^2 = 1\)- Imaginary part: \(2ab = 1\)
04
Express One Variable in Terms of the Other
Use the equation from the imaginary part to express \(b\) in terms of \(a\):\[b = \frac{1}{2a}\]
05
Substitute and Solve for a
Substitute \(b = \frac{1}{2a}\) into the real part equation:\[a^2 - \left(\frac{1}{2a}\right)^2 = 1\]Simplify to obtain:\[a^4 - a^2 = \frac{1}{4}\]
06
Simplify and Solve the Quadratic Equation
Let \(x = a^2\), which transforms the equation into a quadratic:\[x^2 - x - \frac{1}{4} = 0\]Solve this quadratic using the quadratic formula:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where \(a = 1\), \(b = -1\), \(c = -\frac{1}{4}\).
07
Calculate and Choose the Positive Root
Calculate the roots:\[x = \frac{1 \pm \sqrt{1 + \frac{1}{4} \cdot 4}}{2} = \frac{1 \pm \sqrt{2}}{2}\]Choose the positive root \(x = \frac{1 + \sqrt{2}}{2}\). Thus, \(a^2 = \frac{1 + \sqrt{2}}{2}\) and \(a = \sqrt{\frac{1 + \sqrt{2}}{2}}\).
08
Determine b Using a
Find \(b\) using \(b = \frac{1}{2a}\):\[b = \frac{1}{2 \sqrt{\frac{1 + \sqrt{2}}{2}}}\]\[b = \sqrt{\frac{-1 + \sqrt{2}}{2}}\] (rationalizes directly to this form).
09
Express the Solution
Therefore, \(a = \sqrt{\frac{1 + \sqrt{2}}{2}}\) and \(b = \sqrt{\frac{-1 + \sqrt{2}}{2}}\), which confirms the original statement in the question:\[\sqrt{1+i} = \sqrt{\frac{1}{2} + \frac{1}{2} \sqrt{2}} + i \sqrt{-\frac{1}{2} + \frac{1}{2} \sqrt{2}}\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Formula
The quadratic formula is a powerful tool used to find solutions for quadratic equations of the form \( ax^2 + bx + c = 0 \). This formula is especially useful when the equation does not factor easily. The quadratic formula is expressed as:
- \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
- \( a \), \( b \), and \( c \) are the coefficients of the quadratic equation.
- "\( \pm \)" indicates that there are generally two solutions: one with the plus and one with the minus.
- The discriminant, \( \Delta = b^2 - 4ac \), which tells us the nature of the roots (real or complex).
- Substituting into the quadratic formula gives the roots, and choosing the positive root as a balance between simplicity and appropriateness for real-world applications.
Imaginary Unit
The imaginary unit, denoted by \( i \), is a fundamental concept in complex numbers. It is defined as the square root of \(-1\):
- \( i^2 = -1 \)
- \( a \) is the real part.
- \( bi \) is the imaginary part.
Real and Imaginary Parts
In the realm of complex numbers, every number is composed of a real part and an imaginary part. For a complex number \( z = a + bi \):
- The real part is \( a \).
- The imaginary part is \( b \).
- The real part operates like a typical real number and can be plotted on the horizontal axis of the complex plane.
- The imaginary part involves the imaginary unit \( i \), plotted on the vertical axis of the complex plane.
Square Roots of Complex Numbers
Understanding square roots of complex numbers can be fascinating yet challenging. Given a complex number \( z = x + yi \), finding \( \sqrt{z} \) involves ensuring both the real and imaginary parts balance to satisfy the equation when squared. For \( \sqrt{1+i} \), you assume:
- \( \sqrt{1+i} = a + bi \)
- \( a^2 - b^2 = 1 \)
- \( 2ab = 1 \)