Both square roots of 15-8i

Question

I have to find both square roots of 15-8i. I'm guessing it needs to be in a+bi form in the end 

Answer 1

Another solution...

(a+bi)²=15-8i

a²+2abi-b²=15-8i

a²-b²+2abi=15-8i

a²-b²=15 and 2ab=-8

2ab=-8, ab=-4, and b=-4/a

substitute b=-4/a into a²-b²=15

a²-(-4/a)²=15

a²-(16/a²)=15

multiply both sides by a²

a⁴-16=15a²

a⁴-15a²-16=0

let x=a²

x²-15x-16=0

using the quadratic equation

(15+√(225+64))/2

(15+√(289))/2

(15+17)/2

(15+17)/2

32/2=16

x=16, but x=a²

a²=16

+a=4, a=4 and a=-4

b=-4/a

b=-4/4=-1

a=4 and b=-1 giving (4-i)

b=-4/-4=1

a=-4 and b=1 giving (-4+i)

√289 happens to be a perfect square; it would have been more complicated if it wasn't

 

Answer 2

Sarah, the easy way to do this problem is to convert the complex number to its Euler form as Ke^iθ, and then take the square root in which the final answer would be √K *e^i(θ/2) = √K *(cos(θ/2) + i sin(θ/2))

 

K=√(15^2+(-8)^2)=√(225+64)=√289 =17

 

θ=arctan(-8/15) Note: draw a right triangle contain the angle θ with sides 8, 15, and 17. Also, the angle would be in the forth quadrant. 

 

cos(θ)=15/17

sin(θ) = -8/17

 

Use the half angle formulas to determine cos(θ/2) and sin(θ/2).

 

cos(θ/2)=-√((1+cos(θ))/2)=-√((1+15/17)/2) = -√(32/34)=-√(16/17) =-4/√17 Note: The negative answer for cos(θ/2) is due to θ/2 is in the second quadrant and the cosine of the angle becomes negative.

 

sin(θ/2)=√((1-cos(θ))/2) = √((1-15/17)/2) = √(2/17/2) = √(1/17) = 1/√17

 

The answers of the square roots become

 

√17*(-4/√17 + i 1/√17) = -4+i and -√17*(-4/√17+ i 1/√17)= 4-i

 

Test your answers

(-4+i)^2 = 16-8i+i^2=16-8i-1 = 15-8i

(4-i)^2 = 16-8i+i^2 = 16-8i-1 = 15-8i   This checks.