Question

Let z = re^j theta. Express in polar form (i.e., determine the magnitude and angle for) the following functions of z: (i) z* (ii) z^2 (iii) jz (iv) zz* (v) z/z* (vi) 1/z

Answer 1

Answer:

1. |z*| = r   ,   ∠(z*) = -∅
2. |z²| = r²   ,   ∠(z²) = 2∅
3. |jz| = r   ,   ∠(jz) = ∅
4. |zz*| = r²   ,   ∠(zz*) = 0
5. |z/z*| = 1   ,   ∠(z/z*) = 2∅
6. |1/z| = r ⁻¹  ,   ∠(1/z) = -∅

Explanation:

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1.) The * mean you take the conjugate of the value. This means you change the sign of the imaginary part, so if it's positive, turn it negative and vice versa.

2.) The magnitude with an exponent can have the exponent moved outside the magnitude. |zⁿ| = |z|ⁿ

The angle multiplies with its exponent instead. ∠zⁿ = n∠z

3.) This part is just testing if you can convert the number using the eulers formula and convert back.

The magnitude could be found using the distance formula. √(R² + I²)

The angle could be found using tan⁻¹(Imaginary/Real).

4.) Magnitude of a product could be split up. |zv| = |z|·|v|

Angle of a product could be splitted up and added. ∠(zv) = ∠z + ∠v

5.) Simplify it first using some algebra and use the euler's identity to identify the magnitude and angle. It takes in a form like this:

$z=Ae^{j\theta}$

A is your magnitude, ∅ is your angle.

6.) Same rule as part 2

Answer 2

The polar form of the given functions are as follows:

• $z^*=r(sin\theta-jcos\theta)$

• $z^2=r^2(sin2\theta-jcos2\theta)$

• $jz=r \left(sin \left(\dfrac{\pi}{4}+\theta\right)+jcos \left(\dfrac{\pi}{4}+\theta\right)\right)$

• $\dfrac{z}{z^*}=sin2\theta+jcos2\theta$

• $\dfrac{1}{z}=\dfrac{1}{r}(sin\theta-cos\theta)$

In complex number notation:

$z$ represents a complex number

$z^*$ represents the conjugate of the complex number $z$, such that

$z=re^{j\theta}\\z^*=re^{-j\theta}$

in exponential form, or

$z=r(sin\theta+jcos\theta)\\z^*=r(sin\theta-jcos\theta)$

in polar form

Solutions

To express each function in polar form, first simplify, before getting the polar form

(i)    For the complex conjugate $z^*$, we have

$z^*=re^{-j\theta}=r(sin\theta-jcos\theta)$

(ii)  For $z^2$, we have to multiply $z$ by itself, and then use the law of indices to handle the exponential.

$z^2=r\cdot e^{j\theta}\cdot r \cdot e^{j\theta}\\=r\cdot r \cdot e^{(j\theta+j\theta)}\\=r^2e^{j(2\theta)}$

So that we now have

$z^2=r^2(sin2\theta-jcos2\theta)$

(iii)  For $jz$, note that this is a multiplication problem and for the imaginary unit $j$

$modulus(j)=1,arg(j)=\frac{\pi}{4}$

simplifying,

$jz=1\cdot e^{j\frac{\pi}{4}}\cdot r \cdot e^{j\theta}\\=r \cdot e^{(j\frac{\pi}{4}+j\theta)}\\=re^{j(\frac{\pi}{4}+\theta)}$

we can now easily obtain the polar form

$jz=r \left(sin \left(\dfrac{\pi}{4}+\theta\right)+jcos \left(\dfrac{\pi}{4}+\theta\right)\right)$

(iv)   To divide a complex by another in polar form, divide their modulus, and subtract their arguments. In thus case

$\dfrac{z}{z^*}=\dfrac{re^{j\theta}}{re^{-j\theta}}\\\\=e^{j(\theta-(-\theta))}\\=e^{j(2\theta)}$

The polar form is

$\dfrac{z}{z^*}=sin2\theta+jcos2\theta$

(v)    Finally, for the reciprocal of the complex number $z$, note that the numerator has a modulus of 1, and an argument of 0.

$\dfrac{1}{z}=\dfrac{1\cdot e^{j0}}{r\cdot e^{j\theta}}\\\\=\dfrac{1}{r}\cdot e^{j0-j\theta}\\\\=\dfrac{1}{r}\cdot e^{-j\theta}$

The polar form is

$\dfrac{1}{z}=\dfrac{1}{r}(sin\theta-cos\theta)$

Learn more about complex numbers in polar form https://brainly.com/question/917768