Given <em>z</em> = 3 + <em>i</em>, right away we can find
(a) square
<em>z</em> ² = (3 + <em>i </em>)² = 3² + 6<em>i</em> + <em>i</em> ² = 9 + 6<em>i</em> - 1 = 8 + 6<em>i</em>
(b) modulus
|<em>z</em>| = √(3² + 1²) = √(9 + 1) = √10
(d) polar form
First find the argument:
arg(<em>z</em>) = arctan(1/3)
Then
<em>z</em> = |<em>z</em>| exp(<em>i</em> arg(<em>z</em>))
<em>z</em> = √10 exp(<em>i</em> arctan(1/3))
or
<em>z</em> = √10 (cos(arctan(1/3)) + <em>i</em> sin(arctan(1/3))
(c) square root
Any complex number has 2 square roots. Using the polar form from part (d), we have
√<em>z</em> = √(√10) exp(<em>i</em> arctan(1/3) / 2)
and
√<em>z</em> = √(√10) exp(<em>i</em> (arctan(1/3) + 2<em>π</em>) / 2)
Then in standard rectangular form, we have
![\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right)\right)](https://tex.z-dn.net/?f=%5Csqrt%20z%20%3D%20%5Csqrt%5B4%5D%7B10%7D%20%5Cleft%28%5Ccos%5Cleft%28%5Cdfrac12%20%5Carctan%5Cleft%28%5Cdfrac13%5Cright%29%5Cright%29%20%2B%20i%20%5Csin%5Cleft%28%5Cdfrac12%20%5Carctan%5Cleft%28%5Cdfrac13%5Cright%29%5Cright%29%5Cright%29)
and
![\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right)\right)](https://tex.z-dn.net/?f=%5Csqrt%20z%20%3D%20%5Csqrt%5B4%5D%7B10%7D%20%5Cleft%28%5Ccos%5Cleft%28%5Cdfrac12%20%5Carctan%5Cleft%28%5Cdfrac13%5Cright%29%20%2B%20%5Cpi%5Cright%29%20%2B%20i%20%5Csin%5Cleft%28%5Cdfrac12%20%5Carctan%5Cleft%28%5Cdfrac13%5Cright%29%20%2B%20%5Cpi%5Cright%29%5Cright%29)
We can simplify this further. We know that <em>z</em> lies in the first quadrant, so
0 < arg(<em>z</em>) = arctan(1/3) < <em>π</em>/2
which means
0 < 1/2 arctan(1/3) < <em>π</em>/4
Then both cos(1/2 arctan(1/3)) and sin(1/2 arctan(1/3)) are positive. Using the half-angle identity, we then have
![\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}](https://tex.z-dn.net/?f=%5Ccos%5Cleft%28%5Cdfrac12%20%5Carctan%5Cleft%28%5Cdfrac13%5Cright%29%5Cright%29%20%3D%20%5Csqrt%7B%5Cdfrac%7B1%2B%5Ccos%5Cleft%28%5Carctan%5Cleft%28%5Cdfrac13%5Cright%29%5Cright%29%7D2%7D)
![\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}](https://tex.z-dn.net/?f=%5Csin%5Cleft%28%5Cdfrac12%20%5Carctan%5Cleft%28%5Cdfrac13%5Cright%29%5Cright%29%20%3D%20%5Csqrt%7B%5Cdfrac%7B1-%5Ccos%5Cleft%28%5Carctan%5Cleft%28%5Cdfrac13%5Cright%29%5Cright%29%7D2%7D)
and since cos(<em>x</em> + <em>π</em>) = -cos(<em>x</em>) and sin(<em>x</em> + <em>π</em>) = -sin(<em>x</em>),
![\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}](https://tex.z-dn.net/?f=%5Ccos%5Cleft%28%5Cdfrac12%20%5Carctan%5Cleft%28%5Cdfrac13%5Cright%29%2B%5Cpi%5Cright%29%20%3D%20-%5Csqrt%7B%5Cdfrac%7B1%2B%5Ccos%5Cleft%28%5Carctan%5Cleft%28%5Cdfrac13%5Cright%29%5Cright%29%7D2%7D)
![\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}](https://tex.z-dn.net/?f=%5Csin%5Cleft%28%5Cdfrac12%20%5Carctan%5Cleft%28%5Cdfrac13%5Cright%29%2B%5Cpi%5Cright%29%20%3D%20-%5Csqrt%7B%5Cdfrac%7B1-%5Ccos%5Cleft%28%5Carctan%5Cleft%28%5Cdfrac13%5Cright%29%5Cright%29%7D2%7D)
Now, arctan(1/3) is an angle <em>y</em> such that tan(<em>y</em>) = 1/3. In a right triangle satisfying this relation, we would see that cos(<em>y</em>) = 3/√10 and sin(<em>y</em>) = 1/√10. Then
![\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10+3\sqrt{10}}{20}}](https://tex.z-dn.net/?f=%5Ccos%5Cleft%28%5Cdfrac12%20%5Carctan%5Cleft%28%5Cdfrac13%5Cright%29%5Cright%29%20%3D%20%5Csqrt%7B%5Cdfrac%7B1%2B%5Cdfrac3%7B%5Csqrt%7B10%7D%7D%7D2%7D%20%3D%20%5Csqrt%7B%5Cdfrac%7B10%2B3%5Csqrt%7B10%7D%7D%7B20%7D%7D)
![\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10-3\sqrt{10}}{20}}](https://tex.z-dn.net/?f=%5Csin%5Cleft%28%5Cdfrac12%20%5Carctan%5Cleft%28%5Cdfrac13%5Cright%29%5Cright%29%20%3D%20%5Csqrt%7B%5Cdfrac%7B1-%5Cdfrac3%7B%5Csqrt%7B10%7D%7D%7D2%7D%20%3D%20%5Csqrt%7B%5Cdfrac%7B10-3%5Csqrt%7B10%7D%7D%7B20%7D%7D)
![\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}](https://tex.z-dn.net/?f=%5Ccos%5Cleft%28%5Cdfrac12%20%5Carctan%5Cleft%28%5Cdfrac13%5Cright%29%2B%5Cpi%5Cright%29%20%3D%20-%5Csqrt%7B%5Cdfrac%7B10-3%5Csqrt%7B10%7D%7D%7B20%7D%7D)
![\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}](https://tex.z-dn.net/?f=%5Csin%5Cleft%28%5Cdfrac12%20%5Carctan%5Cleft%28%5Cdfrac13%5Cright%29%2B%5Cpi%5Cright%29%20%3D%20-%5Csqrt%7B%5Cdfrac%7B10-3%5Csqrt%7B10%7D%7D%7B20%7D%7D)
So the two square roots of <em>z</em> are
![\boxed{\sqrt z = \sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Csqrt%20z%20%3D%20%5Csqrt%5B4%5D%7B10%7D%20%5Cleft%28%5Csqrt%7B%5Cdfrac%7B10%2B3%5Csqrt%7B10%7D%7D%7B20%7D%7D%20%2B%20i%20%5Csqrt%7B%5Cdfrac%7B10-3%5Csqrt%7B10%7D%7D%7B20%7D%7D%5Cright%29%7D)
and
![\boxed{\sqrt z = -\sqrt[4]{10} \left(\sqrt{\dfrac{10+3\sqrt{10}}{20}} + i \sqrt{\dfrac{10-3\sqrt{10}}{20}}\right)}](https://tex.z-dn.net/?f=%5Cboxed%7B%5Csqrt%20z%20%3D%20-%5Csqrt%5B4%5D%7B10%7D%20%5Cleft%28%5Csqrt%7B%5Cdfrac%7B10%2B3%5Csqrt%7B10%7D%7D%7B20%7D%7D%20%2B%20i%20%5Csqrt%7B%5Cdfrac%7B10-3%5Csqrt%7B10%7D%7D%7B20%7D%7D%5Cright%29%7D)