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Chapter 7: Problem 1

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

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

Step by step solution

01

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.
02

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.
03

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.

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