Question

Write the following in standard form. a. \((4+5 i)(2-3 i)\) b. \((1+i)^{3}\) c. \(\frac{5+3 i}{1-i}\).

Short answer

The answers are: \n(a) -7 + 22i \n(b) -2 + 2i \n(c) 4 + 2i.

Step-by-step solution

Multiply

For (a), multiply the given complex numbers as per the rule of multiplication of complex numbers. (a+bj)(c+dj) = ac-bd + j(ad+bc). Thus, \((4+5i)(2-3i)\) = 4*2 - 5*3 + i(4*3 + 5*2) = -7 + 22i.

Exponentiate

For (b), calculate the cube of the complex number as (a+bj)^3 = a^3 + 3a^2bj + 3ab^2j^2 + b^3j^3. Given number is \(1+i\), so (1+i)^3 = 1 + 3*1*1i + 3*1*1*(-1) + 1*(-1) = -2 + 2i.

Divide

For (c), divide the complex numbers as per the rule of division of complex numbers. \(\frac{a+bj}{c+dj}\) = \(\frac{ac+bd}{c^{2}+d^{2}}\) + j\(\frac{bc-ad}{c^{2}+d^{2}}\). Thus, \(\frac{5+3i}{1-i}\) = \(\frac{5*1 + 3*1}{1^2 + 1^2}\) + j\(\frac{3*1 - 5(-1)}{1^2 + 1^2}\) = 4 + 2i.

Key concepts

Complex Multiplication

Multiplying complex numbers is just like distributing terms in algebra. We use the formula

• For two complex numbers \((a + bi)(c + di)\), the multiplication is done as:
• \(ac - bd + i(ad + bc)\).

Let's look at an example using this formula:

• For \((4 + 5i)(2 - 3i)\):
• Multiply the real parts: \(4 \times 2 = 8\).
• Multiply the imaginary parts: \(-5 \times -3 = 15\).
• For the cross terms: \((4 \times -3) + (5 \times 2) = -12 + 10\).
• Combine the results to get \(-7 + 22i\).

This is how complex numbers are multiplied. It involves distributing just like you do with variables, combining terms, and remembering that \(i^2 = -1\).
By learning these steps, you can multiply any pair of complex numbers easily.

Complex Exponentiation

Exponentiating complex numbers means raising them to a power. We use the formula for cubes:

• For \((a + bi)^3\):
• Use \(a^3 + 3a^2bi + 3ab^2i^2 + b^3i^3\).

Here's an example with \((1+i)^3\):

• The first part: \(1^3 = 1\).
• The second part: \(3 \times 1^2 \times i = 3i\).
• The third part: \(3 \times 1 \times (-1) = -3\) because \(i^2 = -1\).
• The fourth part: \((-1)\) because \(i^3 = i(i^2) = -i\).
• Combine everything to get \(-2 + 2i\).

Complex numbers can often give surprising results when exponentiated. This step-by-step method makes it simpler to handle and understand these seemingly complex processes.

Complex Division

Dividing complex numbers involves multiplying by the conjugate of the denominator. The conjugate of a complex number \(c + di\) is \(c - di\).

• For \(\frac{a + bi}{c + di}\), multiply both numerator and denominator by the conjugate \(c - di\).
• This removes the imaginary part from the denominator:
• \(\frac{(a+bi)(c-di)}{(c+di)(c-di)}\).
• The denominator becomes real: \(c^2 + d^2\).

For example, \(\frac{5 + 3i}{1 - i}\):

• Multiply both parts by \(1 + i\):
• \((5 + 3i)(1 + i) \) in the numerator.
• \(1^2 + 1^2 = 2\) in the denominator.
• Compute: \((5 \times 1 - 3 \times 1) + i(5 \times 1 + 3 \times 1) = 8 + 8i\).
• Result: \(\frac{8}{2} + \frac{8i}{2} = 4 + 2i\).

So, dividing complex numbers is straightforward when using conjugates. It simplifies the division process by turning the denominator into a real number.