calculate \\sqrt{-1 - 3i}. give your answer in a + bi form. give the solution with smallest positive angle.

Explanation:

Step1: Convert to Polar Form

Let ( z = -1 - 3i ). First, find the modulus ( r ) and argument ( \theta ).
The modulus ( r = \sqrt{(-1)^2 + (-3)^2} = \sqrt{1 + 9} = \sqrt{10} ).
The argument ( \theta ) is in the third quadrant (since both real and imaginary parts are negative).
( \tan\theta = \frac{-3}{-1} = 3 ), so ( \theta = \pi + \arctan(3) ).

Step2: Apply De Moivre's Theorem

To find ( \sqrt{z} ), we use De Moivre's Theorem: ( \sqrt[n]{z} = r^{1/n} \left( \cos\left( \frac{\theta + 2k\pi}{n} \right) + i\sin\left( \frac{\theta + 2k\pi}{n} \right) \right) ), where ( n = 2 ) and ( k = 0, 1 ).

For ( k = 0 ):
( r^{1/2} = \sqrt{\sqrt{10}} = 10^{1/4} \approx \sqrt{3.1623} \approx 1.778 ) (but we'll keep it exact for now).
( \frac{\theta}{2} = \frac{\pi + \arctan(3)}{2} = \frac{\pi}{2} + \frac{\arctan(3)}{2} ). Wait, no—correctly, ( \theta = \pi + \arctan(3) ), so ( \frac{\theta}{2} = \frac{\pi}{2} + \frac{\arctan(3)}{2} )? No, wait: ( \theta = \pi + \alpha ), where ( \alpha = \arctan(3) ), so ( \frac{\theta}{2} = \frac{\pi}{2} + \frac{\alpha}{2} ). But actually, let's compute ( \cos\left( \frac{\theta}{2} \right) ) and ( \sin\left( \frac{\theta}{2} \right) ) using half-angle formulas.

Alternatively, let ( \sqrt{-1 - 3i} = a + bi ), then ( (a + bi)^2 = -1 - 3i ).
Expanding: ( a^2 - b^2 + 2abi = -1 - 3i ).
So we have the system:

1. ( a^2 - b^2 = -1 )
2. ( 2ab = -3 )

From equation 2: ( b = \frac{-3}{2a} ). Substitute into equation 1:
( a^2 - \left( \frac{9}{4a^2} \right) = -1 )
Multiply by ( 4a^2 ): ( 4a^4 - 9 = -4a^2 )
( 4a^4 + 4a^2 - 9 = 0 )
Let ( u = a^2 ): ( 4u^2 + 4u - 9 = 0 )
Solve using quadratic formula: ( u = \frac{-4 \pm \sqrt{16 + 144}}{8} = \frac{-4 \pm \sqrt{160}}{8} = \frac{-4 \pm 4\sqrt{10}}{8} = \frac{-1 \pm \sqrt{10}}{2} )

Since ( a^2 ) must be positive, ( u = \frac{-1 + \sqrt{10}}{2} ) (because ( \frac{-1 - \sqrt{10}}{2} ) is negative). Thus, ( a = \pm \sqrt{\frac{-1 + \sqrt{10}}{2}} ).

From ( 2ab = -3 ), if ( a ) is positive, ( b ) is negative; if ( a ) is negative, ( b ) is positive. We need the solution with the smallest positive angle. The angle of ( a + bi ) is ( \arctan\left( \frac{b}{a} \right) ). Let's check signs:

If ( a > 0 ), then ( b = \frac{-3}{2a} < 0 ), so the angle is in the fourth quadrant.
If ( a < 0 ), then ( b = \frac{-3}{2a} > 0 ), angle in the second quadrant.

The smallest positive angle is in the fourth quadrant (since the original complex number ( -1 - 3i ) is in the third quadrant, its square roots should be in the second and fourth quadrants; the fourth quadrant has a smaller positive angle than the second). So we take ( a > 0 ), ( b < 0 ).

Compute ( a = \sqrt{\frac{-1 + \sqrt{10}}{2}} \approx \sqrt{\frac{-1 + 3.1623}{2}} \approx \sqrt{\frac{2.1623}{2}} \approx \sqrt{1.0811} \approx 1.04 )
( b = \frac{-3}{2a} \approx \frac{-3}{2 \times 1.04} \approx \frac{-3}{2.08} \approx -1.442 )

Wait, but let's solve the system exactly. Let ( a^2 - b^2 = -1 ) and ( 2ab = -3 ). Let ( b = - \frac{3}{2a} ), substitute into first equation:

( a^2 - \left( \frac{9}{4a^2} \right) = -1 )
Multiply by ( 4a^2 ): ( 4a^4 + 4a^2 - 9 = 0 )
Let ( u = a^2 ), so ( 4u^2 + 4u - 9 = 0 )
Solutions: ( u = \frac{ -4 \pm \sqrt{16 + 144} }{8} = \frac{ -4 \pm \sqrt{160} }{8} = \frac{ -4 \pm 4\sqrt{10} }{8} = \frac{ -1 \pm \sqrt{10} }{2} )

Since ( u = a^2 \geq 0 ), we take ( u = \frac{ -1 + \sqrt{10} }{2} ) (because ( \sqrt{10} \approx 3.16 > 1 ), so numerator positive). Thus, ( a = \sqrt{ \frac{ \sqrt{10} - 1 }{2} } ), ( b = - \frac{3}{2a} = - \frac{3}{2 \sqrt{ \frac{ \sqrt{10} - 1 }{2} }} = - \frac{3 \sqrt{2} }{2 \sqrt{ \sqrt{10} - 1 }} ). Rationalizing the denominator: multiply numerator and denominator by ( \sqrt{ \sqrt{10} + 1 } ):

( b = - \frac{3 \sqrt{2} \sqrt{ \sqrt{10} + 1 } }{2 \sqrt{ (\sqrt{10} - 1)(\sqrt{10} + 1) }} = - \frac{3 \sqrt{2} \sqrt{ \sqrt{10} + 1 } }{2 \sqrt{9}} = - \frac{3 \sqrt{2} \sqrt{ \sqrt{10} + 1 } }{6} = - \frac{ \sqrt{2} \sqrt{ \sqrt{10} + 1 } }{2} )

But this is complicated. Alternatively, compute numerically:

( \sqrt{10} \approx 3.16227766 )
( \frac{ \sqrt{10} - 1 }{2} \approx \frac{2.16227766}{2} \approx 1.08113883 )
( a = \sqrt{1.08113883} \approx 1.04 ) (wait, no: ( \sqrt{1.0811} \approx 1.04 )? Wait, ( 1.04^2 = 1.0816 ), which is very close. So ( a \approx 1.04 ), ( b = -3/(2*1.04) \approx -1.442 ). Wait, but let's check ( (1.04 - 1.442i)^2 ):

( (1.04)^2 - (1.442)^2 + 21.04(-1.442)i )
( 1.0816 - 2.080 + (-3.00)i )
( -0.9984 - 3.00i \approx -1 - 3i ), which matches. So the approximate value is ( 1.04 - 1.44i ), but let's get the exact form or better approximation.

Wait, actually, let's use the half-angle formulas. The original complex number ( z = -1 - 3i ) has ( r = \sqrt{10} ), ( \theta = \pi + \arctan(3) ). So the square roots are ( \sqrt{r} \left( \cos\left( \frac{\theta}{2} \right) + i\sin\left( \frac{\theta}{2} \right) \right) ) and ( \sqrt{r} \left( \cos\left( \frac{\theta}{2} + \pi \right) + i\sin\left( \frac{\theta}{2} + \pi \right) \right) ).

( \frac{\theta}{2} = \frac{\pi + \arctan(3)}{2} = \frac{\pi}{2} + \frac{\arctan(3)}{2} )? No, ( \theta = \pi + \alpha ), so ( \frac{\theta}{2} = \frac{\pi}{2} + \frac{\alpha}{2} ), where ( \alpha = \arctan(3) ). But ( \cos\left( \frac{\pi}{2} + \frac{\alpha}{2} \right) = -\sin\left( \frac{\alpha}{2} \right) ), ( \sin\left( \frac{\pi}{2} + \frac{\alpha}{2} \right) = \cos\left( \frac{\alpha}{2} \right) ). Wait, no—if ( \theta = \pi + \alpha ), then ( \frac{\theta}{2} = \frac{\pi}{2} + \frac{\alpha}{2} ), so:

( \cos\left( \frac{\theta}{2} \right) = \cos\left( \frac{\pi}{2} + \frac{\alpha}{2} \right) = -\sin\left( \frac{\alpha}{2} \right) )
( \sin\left( \frac{\theta}{2} \right) = \sin\left( \frac{\pi}{2} + \frac{\alpha}{2} \right) = \cos\left( \frac{\alpha}{2} \right) )

But ( \alpha = \arctan(3) ), so ( \sin\left( \frac{\alpha}{2} \right) = \sqrt{ \frac{1 - \cos\alpha}{2} } ), ( \cos\left( \frac{\alpha}{2} \right) = \sqrt{ \frac{1 + \cos\alpha}{2} } ).

Since ( \tan\alpha = 3 ), we can imagine a right triangle with opposite 3, adjacent 1, hypotenuse ( \sqrt{10} ). Thus, ( \cos\alpha = \frac{1}{\sqrt{10}} ), ( \sin\alpha = \frac{3}{\sqrt{10}} ).

Thus, ( \sin\left( \frac{\alpha}{2} \right) = \sqrt{ \frac{1 - \frac{1}{\sqrt{10}} }{2} } ), ( \cos\left( \frac{\alpha}{2} \right) = \sqrt{ \frac{1 + \frac{1}{\sqrt{10}} }{2} } ).

Therefore, for ( k = 0 ):
( \cos\left( \frac{\theta}{2} \right) = -\sin\left( \frac{\alpha}{2} \right) = - \sqrt{ \frac{1 - \frac{1}{\sqrt{10}} }{2} } )
( \sin\left( \frac{\theta}{2} \right) = \cos\left( \frac{\alpha}{2} \right) = \sqrt{ \frac{1 + \frac{1}{\sqrt{10}} }{2} } )

Multiply by ( \sqrt{r} = \sqrt{\sqrt{10}} = 10^{1/4} ):

( \sqrt{z} = 10^{1/4} \left( - \sqrt{ \frac{1 - \frac{1}{\sqrt{10}} }{2} } + i \sqrt{ \frac{1 + \frac{1}{\sqrt{10}} }{2} } \right) ) (this is the second quadrant root)

For ( k = 1 ):
( \frac{\theta + 2\pi}{2} = \frac{\pi + \arctan(3) + 2\pi}{2} = \frac{3\pi + \arctan(3)}{2} = \pi + \frac{\pi + \arctan(3)}{2} ), so:
( \cos\left( \frac{\theta + 2\pi}{2} \right) = \cos\left( \pi + \frac{\theta}{2} \right) = -\cos\left( \frac{\theta}{2} \right) = \sin\left( \frac{\alpha}{2} \right) )
( \sin\left( \frac{\theta + 2\pi}{2} \right) = \sin\left( \pi + \frac{\theta}{2} \right) = -\sin\left( \frac{\theta}{2} \right) = -\cos\left( \frac{\alpha}{2} \right) )

Thus, the root for ( k = 1 ) is:
( \sqrt{z} = 10^{1/4} \left( \sqrt{ \frac{1 - \frac{1}{\sqrt{10}} }{2} } - i \sqrt{ \frac{1 + \frac{1}{\sqrt{10}} }{2} } \right) ) (fourth quadrant root, smallest positive angle)

Now, compute numerically:

( 10^{1/4} \approx \sqrt{3.1623} \approx 1.778 )
( \frac{1 - \frac{1}{\sqrt{10}}}{2} \approx \frac{1 - 0.3162}{2} \approx \frac{0.6838}{2} \approx 0.3419 ), ( \sqrt{0.3419} \approx 0.5848 )
( \frac{1 + \frac{1}{\sqrt{10}}}{2} \approx \frac{1 + 0.3162}{2} \approx \frac{1.3162}{2} \approx 0.6581 ), ( \sqrt{0.6581} \approx 0.8112 )

Thus, for ( k = 1 ):
( \cos\left( \frac{\theta + 2\pi}{2} \right) \approx 1.778 \times 0.5848 \approx 1.04 )
( \sin\left( \frac{\theta + 2\pi}{2} \right) \approx -1.778 \times 0.8112 \approx -1.442 )

So the square root with the smallest positive angle (fourth quadrant) is approximately ( 1.04 - 1.44i ). To get a more accurate value, let's solve ( a^2 - b^2 = -1 ) and ( 2ab = -3 ) exactly. Let ( a = \frac{\sqrt{10} - 1}{2b} )? No, better to use the quadratic solution. Wait, we had ( 4a^4 + 4a^2 - 9 = 0 ), so ( a^2 = \frac{ -1 + \sqrt{10} }{2} \approx \frac{ -1 + 3.1623 }{2} \approx 1.0811 ), so ( a \approx \sqrt{1.0811} \approx 1.04 ), and ( b = -3/(2a) \approx -3/(2*1.04) \approx -1.442 ).

Let's check ( (1.04 - 1.44i)^2 ):
( (1.04)^2 = 1.0816 ), ( (1.44i)^2 = -2.0736 ), cross term ( 21.04(-1.44i) = -3.00i )
So total: ( 1.081