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nevsk [136]
3 years ago
11

Let z=3+i, then find a. Z²b. |Z| c.

rt{Z}" alt="\sqrt{Z}" align="absmiddle" class="latex-formula">
d.  Polar form of z​
Mathematics
2 answers:
zysi [14]3 years ago
3 0

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)

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 <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}

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

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}

\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 <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}}

\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 <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)}

and

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

Nataly_w [17]3 years ago
3 0

Answer:

\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}

<u>Part A:</u>

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}

<u>Part B:</u>

The magnitude, or modulus, of some complex number a+bi is given by \sqrt{a^2+b^2}.

In 3+i, assign values:

  • a=3
  • b=1

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

<u>Part C:</u>

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,\\\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}}}

<u>Part D:</u>

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(\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 z 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}))

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