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Chapter 2: Problem 54

Perform the indicated operation(s) and write the result in standard form. $$(4-i)^{2}-(1+2 i)^{2}$$

Short Answer

Expert verified
The result in standard form is \(18-12i\).

Step by step solution

01

Square the first complex number

The first operation to perform is the squaring of the first complex number, \(4-i\). This equals \((4-i)^{2}= (4-i)(4-i) = 16 - 4i - 4i + i^{2} =16 - 8i +i^{2}\). Here, remember that \(i^2\) is equal to -1, so replacing it we get \(16-8i -1 = 15 - 8i\).
02

Square the second complex number

The next operation to perform is the squaring of the second complex number, \(1+2i\). This equals \((1+2i)^{2} = (1+2i)(1+2i) = 1 + 2i + 2i +4i^{2} = 1+ 4i + 4i^{2}\). Remembering again that \(i^2\) equals -1, replacing it gives \(1+4i -4 =- 3 +4i\).
03

Subtract the squares

Finally, subtract the square of the second complex number from the square of the first one. This gives \((15 - 8i) - (-3 +4i) = 15 - 8i + 3 - 4i = 18-12i)\.

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