Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 6: Problem 57
Solve each problem. Use the quadratic formula and De Moivre's theorem to solve $$ x^{2}+(-1+i) x-i=0 $$
Short Answer
Expert verified
The two solutions are x= -0.780776 i , x = -0.2192237 i
Step by step solution
01
- Identify coefficients
In the quadratic equation of the form \(ax^2 + bx + c = 0\)identify the coefficients a, b, and c. Here,a = 1b = -1 + ic = -i
02
- Quadratic formula
The quadratic formula is given by:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Use this formula to find the roots of the quadratic equation.
03
- Calculate discriminant
The discriminant is given by:\[ \Delta = b^2 - 4ac \]Substitute the values of a, b, and c to find the discriminant.
04
- Substitute values in discriminant
Substitute a = 1, b = -1 + i, and c = -i into the discriminant formula:\[ \Delta = (-1 + i)^2 - 4(1)(-i) \]Simplify to get:\[ \Delta = 1 - 2i + i^2 + 4i \]Since \(i^2 = -1\), it becomes:\[ \Delta = 1 - 2i - 1 + 4i = 2i \]
05
- Calculate square root of discriminant
To find the square root of \(2i\)convert it to polar form. \(2i = 2e^{i \frac{\pi}{2}}\)Taking the square root using De Moivre's theorem,\[\sqrt{2i} = \sqrt{2}e^{i \frac{\pi}{4}} \]
06
- Apply the quadratic formula
Now, apply the quadratic formula with \(b = -1 + i\) and \(\sqrt{\Delta} = \sqrt{2}e^{i \frac{\pi}{4}}\):\[ x = \frac{-(-1 + i) \pm \sqrt{2}e^{i \frac{\pi}{4}}}{2} \]}, {
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
quadratic formula
To solve a quadratic equation like the one given, we use the quadratic formula. The general form of a quadratic equation is:
- \[ ax^2 + bx + c = 0 \]
- The quadratic formula to find the roots is:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
- \( a = 1 \)
- \( b = -1 + i \)
- \( c = -i \)
complex numbers
Complex numbers are numbers that have both real and imaginary parts. They are generally represented in the form:
- \( z = a + bi \)
- \( a \) is the real part
- \( b \) is the imaginary part
- \( i \) is the imaginary unit and \( i^2 = -1. \)
De Moivre's theorem
De Moivre's theorem helps us find powers and roots of complex numbers when they are in polar form. The theorem states:
- \[ (re^{i\theta})^n = r^n e^{in\theta} \]
- \( 2i = 2e^{i \frac{\pi}{2}} \)
- \[ \sqrt{2i} = \sqrt{2}e^{i \frac{\pi}{4}} \]
discriminant
The discriminant in a quadratic equation helps determine the nature of the roots. It is given by:
- \[ \Delta = b^2 - 4ac \]
- \( a = 1 \)
- \( b = -1 + i \)
- \( c = -i \)
- \( \Delta = (-1 + i)^2 - 4(1)(-i) \)
\( \Delta = (1 - 2i + i^2) + 4i \)
Since \(i^2 = -1\),
\( \Delta = 1 - 2i - 1 + 4i = 2i \)
polar form
Polar form is a way of expressing complex numbers using their magnitude and angle. It is useful in complex number calculations, especially with De Moivre's theorem. The polar form representation of a complex number \( z = a + bi \) is:
- \[ z = re^{i\theta} \]
- \( r = \sqrt{a^2 + b^2} \) (magnitude)
- \( \theta = \arctan\frac{b}{a} \) (angle or argument)
- \( 2i = 2e^{i \frac{\pi}{2}} \)
