Question

Let z=3+i,

then find
a. Z²
b. |Z|
c.
`√(Z)`
d.  Polar form of z​

Answer 1

Given z = 3 + i, right away we can find

(a) square

z ² = (3 + i )² = 3² + 6i + i ² = 9 + 6i - 1 = 8 + 6i

(b) modulus

|z| = √(3² + 1²) = √(9 + 1) = √10

(d) polar form

First find the argument:

arg(z) = arctan(1/3)

Then

z = |z| exp(i arg(z))

z = √10 exp(i arctan(1/3))

or

z = √10 (cos(arctan(1/3)) + i sin(arctan(1/3))

(c) square root

Any complex number has 2 square roots. Using the polar form from part (d), we have

√z = √(√10) exp(i arctan(1/3) / 2)

and

√z = √(√10) exp(i (arctan(1/3) + 2π) / 2)

Then in standard rectangular form, we have

`\sqrt z = \sqrt[4]{10} \left(\cos\left(\frac12 \arctan\left(\frac13\right)\right) + i \sin\left(\frac12 \arctan\left(\frac13\right)\right)\right)`

and

`\sqrt z = \sqrt[4]{10} \left(\cos\left(\frac12 \arctan\left(\frac13\right) + \pi\right) + i \sin\left(\frac12 \arctan\left(\frac13\right) + \pi\right)\right)`

We can simplify this further. We know that z lies in the first quadrant, so

0 < arg(z) = arctan(1/3) < π/2

which means

0 < 1/2 arctan(1/3) < π/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(\frac12 \arctan\left(\frac13\right)\right) = \sqrt{\frac{1+\cos\left(\arctan\left(\frac13\right)\right)}2}`

`\sin\left(\frac12 \arctan\left(\frac13\right)\right) = \sqrt{\frac{1-\cos\left(\arctan\left(\frac13\right)\right)}2}`

and since cos(x + π) = -cos(x) and sin(x + π) = -sin(x),

`\cos\left(\frac12 \arctan\left(\frac13\right)+\pi\right) = -\sqrt{\frac{1+\cos\left(\arctan\left(\frac13\right)\right)}2}`

`\sin\left(\frac12 \arctan\left(\frac13\right)+\pi\right) = -\sqrt{\frac{1-\cos\left(\arctan\left(\frac13\right)\right)}2}`

Now, arctan(1/3) is an angle y such that tan(y) = 1/3. In a right triangle satisfying this relation, we would see that cos(y) = 3/√10 and sin(y) = 1/√10. Then

`\cos\left(\frac12 \arctan\left(\frac13\right)\right) = \sqrt{\frac{1+\frac3{√(10)}}2} = \sqrt{(10+3√(10))/(20)}`

`\sin\left(\frac12 \arctan\left(\frac13\right)\right) = \sqrt{\frac{1-\frac3{√(10)}}2} = \sqrt{(10-3√(10))/(20)}`

`\cos\left(\frac12 \arctan\left(\frac13\right)+\pi\right) = -\sqrt{(10-3√(10))/(20)}`

`\sin\left(\frac12 \arctan\left(\frac13\right)+\pi\right) = -\sqrt{(10-3√(10))/(20)}`

So the two square roots of z are

`\boxed{\sqrt z = \sqrt[4]{10} \left(\sqrt{(10+3√(10))/(20)} + i \sqrt{(10-3√(10))/(20)}\right)}`

and

`\boxed{\sqrt z = -\sqrt[4]{10} \left(\sqrt{(10+3√(10))/(20)} + i \sqrt{(10-3√(10))/(20)}\right)}`

Answer 2

`\displaystyle \text{a. }8+6i\\\\\text{b. }√(10)\\\\\text{c. }\\\sqrt{\sqrt{(5)/(2)}+(3)/(2)}+i\sqrt{(√(10)-3)/(2)},\\-\sqrt{\sqrt{(5)/(2)}+(3)/(2)}-i\sqrt{(√(10)-3)/(2)}\\\\\\\text{d. }\\\text{Exact: }z=√(10)\left(\cos\left(\arctan\left((1)/(3)\right)\right), i\sin\left(\arctan\left((1)/(3)\right)\right)\right),\\\text{Approximated: }z=3.16(\cos(18.4^(\circ)),i\sin(18.4^(\circ)))`

Explanation:

Recall that
`i=√(-1)`

Part A:

We are just squaring a binomial, so the FOIL method works great. Also, recall that
`(a+b)^2=a^2+2ab+b^2`.

`z^2=(3+i)^2,\\z^2=3^2+2(3i)+i^2,\\z^2=9+6i-1,\\z^2=\boxed{8+6i}`

Part B:

The magnitude, or modulus, of some complex number
`a+bi` is given by
`√(a^2+b^2)`.

In
`3+i`, assign values:

• 
  `a=3`
• 
  `b=1`

`|z|=√(3^2+1^2),\\|z|=√(9+1),\\|z|=√(10)`

Part C:

In Part A, notice that when we square a complex number in the form
`a+bi`, our answer is still a complex number in the form

We have:

`(c+di)^2=a+bi`

Expanding, we get:

`c^2+2cdi+(di)^2=a+bi,\\c^2+2cdi+d^2(-1)=a+bi,\\c^2-d^2+2cdi=a+bi`

This is still in the exact same form as
`a+bi` where:

• 
  `c^2-d^2` corresponds with
  `a`
• 
  `2cd` corresponds with
  `b`

Thus, we have the following system of equations:

`\begin{cases}c^2-d^2=3,\\2cd=1\end{cases}`

Divide the second equation by
`2d` to isolate
`c`:

`2cd=1,\\(2cd)/(2d)=(1)/(2d),\\c=(1)/(2d)`

Substitute this into the first equation:

`\left((1)/(2d)\right)^2-d^2=3,\\(1)/(4d^2)-d^2=3,\\1-4d^4=12d^2,\\-4d^4-12d^2+1=0`

This is a quadratic disguise, let
`u=d^2` and solve like a normal quadratic.

Solving yields:

`d=\pm i \sqrt{(3+√(10))/(2)},\\d=\pm \sqrt{\frac{{√(10)-3}}{2}}`

We stipulate
`d\in \mathbb{R}` and therefore
`d=\pm i \sqrt{(3+√(10))/(2)}` is extraneous.

Thus, we have the following cases:

`\begin{cases}c^2-\left(\sqrt{(√(10)-3)/(2)}\right)^2=3\\c^2-\left(-\sqrt{(√(10)-3)/(2)}\right)^2=3\end{cases}\\`

Notice that
`\left(\sqrt{(√(10)-3)/(2)}\right)^2=\left(-\sqrt{(√(10)-3)/(2)}\right)^2`. However, since
`2cd=1`, two solutions will be extraneous and we will have only two roots.

Solving, we have:

`\begin{cases}c^2-\left(\sqrt{(√(10)-3)/(2)}\right)^2=3 \\c^2-\left(-\sqrt{(√(10)-3)/(2)}\right)^2=3\end{cases}\\\\c^2-\sqrt{(5)/(2)}+(3)/(2)=3,\\c=\pm \sqrt{\sqrt{(5)/(2)}+(3)/(2)`

Given the conditions
`c\in \mathbb{R}, d\in \mathbb{R}, 2cd=1`, the solutions to this system of equations are:

`\left(\sqrt{\sqrt{(5)/(2)}+(3)/(2)}, \sqrt{(√(10)-3)/(2)}\right),\\\left(-\sqrt{\sqrt{(5)/(2)}+(3)/(2)},- (√(10)-3)/(2)}\right)`

Therefore, the square roots of
`z=3+i` are:

`√(z)=\boxed{\sqrt{\sqrt{(5)/(2)}+(3)/(2)}+i\sqrt{(√(10)-3)/(2)} },\\√(z)=\boxed{-\sqrt{\sqrt{(5)/(2)}+(3)/(2)}-i\sqrt{(√(10)-3)/(2)}}`

Part D:

The polar form of some complex number
`a+bi` is given by
`z=r(\cos \theta+\sin \theta)i`, where
`r` is the modulus of the complex number (as we found in Part B), and
`\theta=\arctan((b)/(a))` (derive from right triangle in a complex plane).

We already found the value of the modulus/magnitude in Part B to be
`r=√(10)`.

The angular polar coordinate
`\theta` is given by
`\theta=\arctan((b)/(a))` and thus is:

`\theta=\arctan((1)/(3)),\\\theta=18.43494882\approx 18.4^(\circ)`

Therefore, the polar form of
`z` is:

`\displaystyle \text{Exact: }z=√(10)\left(\cos\left(\arctan\left((1)/(3)\right)\right), i\sin\left(\arctan\left((1)/(3)\right)\right)\right),\\\text{Approximated: }z=3.16(\cos(18.4^(\circ)),i\sin(18.4^(\circ)))`