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Chapter 6: Problem 38

Find a complex number whose square equals \(21-20 i\).

Short Answer

Expert verified
The complex numbers whose square equals \(21 - 20i\) are \(2 - 5i\) and \(-2 + 5i\).

Step by step solution

01

Write the unknown complex number and its square

Let's denote the unknown complex number as \(z = x+yi\), where \(x\) and \(y\) are real numbers. The square of this complex number will be \(z^2 = (x+yi)^2\).
02

Expand the squared complex number

We will now expand \(z^2 = (x+yi)^2\) as follows: \[z^2 = x^2 + 2x(yi) + (yi)^2\]
03

Simplify the squared complex number

Recall that \(i^2 = -1\). So, we can simplify the equation as: \[z^2 = x^2 - y^2 + 2xyi\] Now, let's set the real and imaginary parts of the equation equal to the given complex number \(21 - 20i\). We have: \[x^2 - y^2 = 21\] \[2xy = -20\]
04

Solve the system of equations for x and y

To solve the system of equations, we can use the equation \(2xy = -20\) to find \(y\) in terms of \(x\), and substitute it into the equation \(x^2 - y^2 = 21\). From the equation \(2xy = -20\), we get: \[y = -\frac{10}{x}\] Now substitute this expression for \(y\) into the equation \(x^2 - y^2 = 21\): \[x^2 - \left(-\frac{10}{x}\right)^2 = 21\] Multiply both sides of the equation by \(x^2\) to eliminate the fraction: \[x^4 - 100 = 21x^2\] Move all terms to the left-hand side and set the equation equal to zero: \[x^4 - 21x^2 - 100 = 0\]
05

Find x and y values using the quartic equation

Now we have a quartic equation whose roots represent possible values of \(x\). In this case, we can find that two of the possible roots are \(x = 2\) and \(x = -2\). If we use \(x = 2\), we find that: \[y = -\frac{10}{2} = -5\] If we use \(x = -2\), we find that: \[y = -\frac{10}{-2}= 5\]
06

Write the complex numbers

Using the values of \(x\) and \(y\) we found, we can write our two complex numbers as: \[z_1 = 2 - 5i\] \[z_2 = -2 + 5i\] So, the complex numbers whose square equals \(21 - 20i\) are \(2 - 5i\) and \(-2 + 5i\).

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