Mathematics, 13.08.2021 20:50, genyjoannerubiera

Let z=3+i,
then find
a. Z²
b. |Z|
c. $\sqrt{Z}$
d. Polar form of z

Answers: 2

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Answers

answered: ydesoto0513

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))

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

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

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right)\right)$

and

$\sqrt z = \sqrt[4]{10} \left(\cos\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right) + i \sin\left(\dfrac12 \arctan\left(\dfrac13\right) + \pi\right)\right)$

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

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

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

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1+\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{1-\cos\left(\arctan\left(\dfrac13\right)\right)}2}$

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

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1+\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10+3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)\right) = \sqrt{\dfrac{1-\dfrac3{\sqrt{10}}}2} = \sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\cos\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

$\sin\left(\dfrac12 \arctan\left(\dfrac13\right)+\pi\right) = -\sqrt{\dfrac{10-3\sqrt{10}}{20}}$

So the two square roots of z are

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

and

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

answered: Wintersavannah

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

Step-by-step explanation:

Recall that $i=\sqrt{-1}$

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 $\sqrt{a^2+b^2}$ .

In 3+i , assign values:

$|z|=\sqrt{3^2+1^2},\\|z|=\sqrt{9+1},\\|z|=\sqrt{10}$

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:

corresponds with

corresponds with

Thus, we have the following system of equations:

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

Divide the second equation by to isolate :

$2cd=1,\\\frac{2cd}{2d}=\frac{1}{2d},\\c=\frac{1}{2d}$

Substitute this into the first equation:

$\left(\frac{1}{2d}\right)^2-d^2=3,\\\frac{1}{4d^2}-d^2=3,\\1-4d^4=12d^2,\\-4d^4-12d^2+1=0$

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

Solving yields:

$d=\pm i \sqrt{\frac{3+\sqrt{10}}{2}},\\d=\pm \sqrt{\frac{{\sqrt{10}-3}}{2}}$

We stipulate $d\in \mathbb{R}$ and therefore $d=\pm i \sqrt{\frac{3+\sqrt{10}}{2}}$ is extraneous.

Thus, we have the following cases:

$\begin{cases}c^2-\left(\sqrt{\frac{\sqrt{10}-3}{2}}\right)^2=3\\c^2-\left(-\sqrt{\frac{\sqrt{10}-3}{2}}\right)^2=3\end{cases}\\$

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

Solving, we have:

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

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

$\left(\sqrt{\sqrt{\frac{5}{2}}+\frac{3}{2}}, \sqrt{\frac{\sqrt{10}-3}{2}}\right),\\\left(-\sqrt{\sqrt{\frac{5}{2}}+\frac{3}{2}},- \frac{\sqrt{10}-3}{2}}\right)$

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

$\sqrt{z}=\boxed{\sqrt{\sqrt{\frac{5}{2}}+\frac{3}{2}}+i\sqrt{\frac{\sqrt{10}-3}{2}} },\\\sqrt{z}=\boxed{-\sqrt{\sqrt{\frac{5}{2}}+\frac{3}{2}}-i\sqrt{\frac{\sqrt{10}-3}{2}}}$

Part D:

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

We already found the value of the modulus/magnitude in Part B to be $r=\sqrt{10}$ .

The angular polar coordinate $\theta$ is given by $\theta=\arctan(\frac{b}{a})$ and thus is:

$\theta=\arctan(\frac{1}{3}),\\\theta=18.43494882\approx 18.4^{\circ}$

Therefore, the polar form of is:

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