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Chapter 0: Problem 1

Compute the following complex numbers, and express in the form \(x+i y\), where \(x, y\) are real: (a) \((2-i 3)+(-1+i 6)\) (b) \((4+i 2)-(6-i 3)\) (c) \((6-i \sqrt{2})(2+i 4)\) \(1+i\) (d) \(\frac{1}{1-i}\) (e) \(|4-i 5|\) (f) \(\operatorname{Re}(4-i 5)\) (g) \(\operatorname{Im}(6+i 2)\)

Short Answer

Expert verified
(a) 1 + 3i, (b) -2 + 5i, (c) 12 + 4\sqrt{2} + (24 - 2\sqrt{2})i, (d) \frac{1}{2} + \frac{1}{2}i, (e) \sqrt{41}, (f) 4, (g) 2

Step by step solution

01

Addition of Complex Numbers

Begin by adding the two complex numbers in part (a) of the question. The numbers are \((2-i 3)+(-1+i 6)\). The addition is performed component-wise:\[(2-i3) + (-1+i6) = (2 - 1) + (-3 + 6)i = 1 + 3i\]Thus, the result of part (a) is \(1 + 3i\).
02

Subtraction of Complex Numbers

For part (b), perform subtraction for the complex numbers \((4+i 2)-(6-i 3)\). This is carried out as:\[(4+i2) - (6-i3) = (4-6) + (2-(-3))i = -2 + 5i\]Hence, the answer for part (b) is \(-2 + 5i\).
03

Multiplication of Complex Numbers

Multiply the two complex numbers given in part (c): \((6-i \sqrt{2})(2+i 4)\). Use distributive property (FOIL method):\[(6-i\sqrt{2})(2+i4) = 12 + 24i - 2i\sqrt{2} - i^2\sqrt{2}\times 4\]Remembering \(i^2 = -1\), simplify:\[12 + 24i - 2i\sqrt{2} + 4\sqrt{2} = (12 + 4\sqrt{2}) + (24 - 2\sqrt{2})i\]So, the result is \(12 + 4\sqrt{2} + (24 - 2\sqrt{2})i\).
04

Multiplication by a Complex Conjugate

Rationalize the denominator in part (d): \(\frac{1}{1-i}\). Multiply numerator and denominator by the complex conjugate of the denominator:\[\frac{1}{1-i} \times \frac{1+i}{1+i} = \frac{1+i}{1^2 - (-1)^2} = \frac{1+i}{1+1} = \frac{1+i}{2}\]Thus, the solution simplifies to \(\frac{1}{2} + \frac{1}{2}i\).
05

Magnitude of a Complex Number

Find the magnitude of the complex number in part (e): \(|4-i 5|\). Use the formula:\[|a + bi| = \sqrt{a^2 + b^2}\]Substitute the values:\[|4-i5| = \sqrt{4^2 + (-5)^2} = \sqrt{16 + 25} = \sqrt{41}\]
06

Real Part of a Complex Number

To find the real part of \(4 - i 5\) in part (f), note that the real part is simply the coefficient of the real unit:- The real part is \(4\).
07

Imaginary Part of a Complex Number

To find the imaginary part of \(6+i 2\) in part (g), it is the coefficient of the imaginary unit \(i\):- The imaginary part is \(2\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Addition and Subtraction of Complex Numbers
Complex numbers are a blend of real and imaginary numbers, expressed as \(a + bi\). When dealing with addition or subtraction, the process is straightforward.
  • For addition, such as in the problem \((2 - i3) + (-1 + i6)\): Add the real parts together \, \((2 + (-1))\) \, and the imaginary parts together \, \((-3) + 6\). This results in \(1 + 3i\).
  • For subtraction, as in \((4 + i2) - (6 - i3)\): Subtract the real parts \, \((4 - 6)\) \, and the imaginary parts \, \((2 - (-3))\) \, separately, resulting in \(-2 + 5i\).
This method ensures that you're properly managing both components of complex numbers, making addition and subtraction just as manageable as with regular numbers.
Multiplication of Complex Numbers
When multiplying complex numbers, the distributive property, or the FOIL method (First, Outer, Inner, Last), is your best friend. Take for instance the multiplication problem: \((6 - i\sqrt{2})(2 + i4)\).Start by applying the FOIL method:
  • First: \(6 \times 2 = 12\)
  • Outer: \(6 \times i4 = 24i\)
  • Inner: \(-i\sqrt{2} \times 2 = -2i\sqrt{2}\)
  • Last: \(-i\sqrt{2} \times i4 = -4\sqrt{2}\) (since \(i^2 = -1\))
Combine these products: \(12 + 24i - 2i\sqrt{2} + 4\sqrt{2}\). Thus, the simplified result is \(12 + 4\sqrt{2} + (24 - 2\sqrt{2})i\). Remember that \(i^2\), the square of the imaginary unit \(i\), is \(-1\), which often helps to simplify terms in such expressions.
Magnitude of a Complex Number
The magnitude of a complex number, also known as its modulus, is like measuring its "length" in the complex plane.To find this magnitude \(|a + bi|\), use the formula:\[ |a + bi| = \sqrt{a^2 + b^2} \]For the example \(4 - i 5\):
  • Square the real part: \(4^2 = 16\)
  • Square the imaginary part: \((-5)^2 = 25\)
  • Add these squares: \(16 + 25 = 41\)
  • Finally, take the square root: \(\sqrt{41}\)
The result is \(\sqrt{41}\), giving you the magnitude of the complex number. This measure is essential first to understand how far a complex number is from the origin in a two-dimensional real-imaginary space.
Real and Imaginary Parts of Complex Numbers
Understanding the real and imaginary parts of a complex number is crucial for performing various operations with them. A complex number is expressed as \(a + bi\), where \(a\) is known as the real part and \(b\) is the imaginary part.
  • For \(4 - i 5\), the real part is simply \(4\). Ignore the imaginary unit \(i\) when identifying it.
  • For \(6 + i 2\), the imaginary part is \(2\), which is the coefficient in front of the imaginary unit \(i\).
By separating these parts, complex numbers become easier to comprehend and manipulate. This separation allows you to efficiently tally up, subtract, or reconstruct complex numbers without error.

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Most popular questions from this chapter

Let \(f\) be a complex-valued function defined on a disk \(D: \quad|z|0)\) which is differentiable there. Let $$ u(x, y)=(\operatorname{Re} f)(x+i y), \quad v(x, y)=(\operatorname{Im} f)(x+i y) $$ Show that $$ \frac{\partial u}{\partial x}=\frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y}=-\frac{\partial v}{\partial x} $$ for all \(z-x+i y\) in \(D .\) (Hint: If \(z_{0}=x_{0}+i y_{0}\) is in \(D\), let \(0<\left|z-z_{0}\right| \rightarrow 0\), in the definition of \(f^{\prime}\) \(\left(z_{0}\right)\), through \(z\) of the form \(z=x+i y_{0}\), and then of the form \(z=x_{0}+i y\), to obtain $$ \begin{aligned} f^{\prime}\left(z_{0}\right) &=\frac{\partial u}{\partial x}\left(x_{0}, y_{0}\right)+i \frac{\partial v}{\partial x}\left(x_{0}, y_{0}\right) \\ &=\frac{\partial v}{\partial y}\left(x_{0}, y_{0}\right)-i \frac{\partial u}{\partial y}\left(x_{0}, y_{0}\right) \end{aligned} $$ The equations \(\left({ }^{*}\right)\) are called the Cauchy-Riemann equations.)

Prove that $$ \left|z_{1}+z_{2}\right|^{2}+\left|z_{1}-z_{2}\right|^{2}=2\left|z_{1}\right|^{2}+2\left|z_{2}\right|^{2} $$

If \(r\) is a complex number, and $$ p(z)=(z-r)^{n} $$ where \(n\) is a positive integer, show that $$ p(r)=p^{\prime}(r)=\cdots=p^{(n-1)}(r)=0, \quad p^{(n)}(r)=n ! $$

Solve the following system for \(z_{1}, z_{2}\) and \(z_{3}\) : $$ \begin{array}{r} 3 z_{1}+z_{2}-z_{3}=0 \\ 2 z_{1} \quad-z_{3}=1 \\ z_{2}+2 z_{3}=2 \end{array} $$

If for all real \(x\) $$ f(x)=x+i x^{2}, \quad g(x)=\frac{x^{2}}{2} $$ compute: (a) The function \(F\) given by \(F(x)=f(g(x))\) (b) \(F^{\prime}(x)\)
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