Find the square roots of 119 + 120i algebraically. Let √(119 + 120i) = p + qi be the square root of 119 + 120i. Then (√(119 + 120i))^2 = 119 + 120i and (p + qi)^2 = 119 + 120i. a. Expand the left side of this equation. b. Equate the real and imaginary parts, and solve for p and q. c. What are the square roots of 119 + 120i
The Correct Answer and Explanation is:
To find the square roots of the complex number 119+120i119 + 120i119+120i algebraically, let:119+120i=p+qi\sqrt{119 + 120i} = p + qi119+120i=p+qi
Then squaring both sides gives:(p+qi)2=119+120i(p + qi)^2 = 119 + 120i(p+qi)2=119+120i
a. Expand the left side
(p+qi)2=p2+2pqi+(qi)2=p2+2pqi+q2i2(p + qi)^2 = p^2 + 2pqi + (qi)^2 = p^2 + 2pqi + q^2 i^2(p+qi)2=p2+2pqi+(qi)2=p2+2pqi+q2i2
Since i2=−1i^2 = -1i2=−1, we have:(p+qi)2=p2−q2+2pqi(p + qi)^2 = p^2 – q^2 + 2pq i(p+qi)2=p2−q2+2pqi
So:(p+qi)2=(p2−q2)+(2pq)i(p + qi)^2 = (p^2 – q^2) + (2pq)i(p+qi)2=(p2−q2)+(2pq)i
b. Equate real and imaginary parts
We now equate this to 119+120i119 + 120i119+120i:
- Real part: p2−q2=119p^2 – q^2 = 119p2−q2=119
- Imaginary part: 2pq=1202pq = 1202pq=120
From the second equation:pq=60(Equation 1)pq = 60 \quad \text{(Equation 1)}pq=60(Equation 1)
From the first equation:p2−q2=119(Equation 2)p^2 – q^2 = 119 \quad \text{(Equation 2)}p2−q2=119(Equation 2)
Now use substitution. From Equation 1, q=60pq = \frac{60}{p}q=p60. Plug into Equation 2:p2−(60p)2=119p^2 – \left(\frac{60}{p}\right)^2 = 119p2−(p60)2=119p2−3600p2=119p^2 – \frac{3600}{p^2} = 119p2−p23600=119
Multiply both sides by p2p^2p2 to eliminate the denominator:p4−3600=119p2p^4 – 3600 = 119p^2p4−3600=119p2p4−119p2−3600=0p^4 – 119p^2 – 3600 = 0p4−119p2−3600=0
Let x=p2x = p^2x=p2, then:x2−119x−3600=0x^2 – 119x – 3600 = 0x2−119x−3600=0
Solve using the quadratic formula:x=119±1192+4⋅36002x = \frac{119 \pm \sqrt{119^2 + 4 \cdot 3600}}{2}x=2119±1192+4⋅3600x=119±14161+144002=119±285612x = \frac{119 \pm \sqrt{14161 + 14400}}{2} = \frac{119 \pm \sqrt{28561}}{2}x=2119±14161+14400=2119±2856128561=169\sqrt{28561} = 16928561=169x=119±1692⇒x=144 or x=−25x = \frac{119 \pm 169}{2} \Rightarrow x = 144 \text{ or } x = -25x=2119±169⇒x=144 or x=−25
We discard x=−25x = -25x=−25 since x=p2x = p^2x=p2 must be nonnegative.
So p2=144p^2 = 144p2=144, thus p=±12p = \pm12p=±12
Using Equation 1: pq=60pq = 60pq=60
If p=12p = 12p=12, then q=6012=5q = \frac{60}{12} = 5q=1260=5
If p=−12p = -12p=−12, then q=60−12=−5q = \frac{60}{-12} = -5q=−1260=−5
c. Final Answer
The square roots of 119+120i119 + 120i119+120i are:12+5iand−12−5i\boxed{12 + 5i \quad \text{and} \quad -12 – 5i}12+5iand−12−5i
Explanation
To find the square root of a complex number like 119+120i119 + 120i119+120i, we assume the root takes the form of p+qip + qip+qi, where both ppp and qqq are real numbers. Squaring this expression gives a new complex number, which we can compare to the original. Specifically:(p+qi)2=p2−q2+2pqi(p + qi)^2 = p^2 – q^2 + 2pqi(p+qi)2=p2−q2+2pqi
By comparing this to 119+120i119 + 120i119+120i, we match the real part p2−q2=119p^2 – q^2 = 119p2−q2=119 and the imaginary part 2pq=1202pq = 1202pq=120. This gives us a system of equations. We isolate one variable using the second equation: pq=60pq = 60pq=60, so q=60pq = \frac{60}{p}q=p60. Substituting into the first equation yields an equation with one variable:p2−(60p)2=119p^2 – \left(\frac{60}{p}\right)^2 = 119p2−(p60)2=119
To eliminate the fraction, we multiply through by p2p^2p2, leading to a quartic equation. After substitution x=p2x = p^2x=p2, the equation becomes a quadratic in xxx. Solving with the quadratic formula, we find the discriminant is a perfect square, which makes this problem solvable without approximation.
The valid solution from the quadratic is x=144x = 144x=144, so p=±12p = \pm12p=±12. Using pq=60pq = 60pq=60, we find the corresponding q=±5q = \pm5q=±5, giving us two square roots: 12+5i12 + 5i12+5i and −12−5i-12 – 5i−12−5i. These are complex conjugates, as is typical for square roots of nonreal complex numbers.
This algebraic method avoids converting to polar form and is useful when exact values are needed.
