Complex Numbers - a review problem (a - g)

[VinnyCee]

Homework Statement

Evaluate each of the following complex numbers and express the result in rectangular form:

a) $$4\,e^{j\frac{\pi}{3}}$$

b) $$\sqrt{3}\,e^{j\frac{3\pi}{4}}$$

c) $$6\,e^{-j\frac{\pi}{2}}$$

d) $$j^3$$

e) $$j^{-4}$$

f) $$\left(1\,-\,j\right)^2$$

g) $$\left(1\,-\,j\right)^{\frac{1}{2}}$$

Homework Equations

$$e^{j\Theta}\,=\,cos\,\Theta\,+\,j\,sin\,\Theta$$

$$|z|\,=\,+\sqrt{z\,z^*}$$

where $z^*$ is the complex conjugate

stuff like that...

The Attempt at a Solution

a) $$2\,+\,j\,2\,\sqrt{3}$$

b) $$-\frac{\sqrt{3}}{\sqrt{2}}\,+\,j\,\frac{\sqrt{3}}{\sqrt{2}}$$

c) -6 j

d) -j

e) 1

f) $$-2\,-\,2\,j$$

g) $$-\frac{1}{4}\,j$$

Are these correct? I don't think the last one (g) is right...

 

[olgranpappy]

Are these correct? I don't think the last one (g) is right...

g is indeed not correct. It's $2^{(1/4)}e^{-j\pi/8}$.

 

[VinnyCee]

The others are fine though? How do I do the last one then?

 

[HallsofIvy]

Writing a solution does not mean just showing the answers. How did you get those?

 

[olgranpappy]

The others are fine though? How do I do the last one then?

I don't know if the others are fine; i didn't check them.

to do the last one you could write 1-j in "polar coordinates", after that taking the square root is easy. cheers.

 

[VinnyCee]

a) $$4\left[cos\left(\frac{\pi}{3}\right)\,+\,j\,sin\left(\frac{\pi}{3}\right)\right]\,=\,4\left(\frac{1}{2}\,+\,j\,\frac{\sqrt{3}}{2}\right)\,=\,2\,+\,j\,2\,\sqrt{3}$$

b) $$\sqrt{3}\left[cos\left(\frac{3\,\pi}{4}\right)\,+\,j\,sin\left(\frac{3\,\pi}{4}\right)\right]\,=\,\sqrt{3}\left(-\frac{1}{\sqrt{2}}\,+\,j\,\frac{1}{\sqrt{2}}\right)\,=\,-\frac{\sqrt{3}}{2}\,+\,j\,\frac{\sqrt{3}}{\sqrt{2}}$$

c) $$6\left[cos\left(-\frac{\pi}{2}\right)\,+\,j\,sin\left(-\frac{\pi}{2}\right)\right]\,=\,6\left[0\,+\,j\,(-1)\right]\,=\,-6j$$

d) $$j^3\,=\,j^2\,\cdot\,j\,=\,-j$$

e) $$j^{-4}\,=\,\frac{1}{j^4}\,=\,\frac{1}{(-1)^2}\,=\,1$$

f) $$\left(1\,-\,j\right)^3\,=\,\left(1\,-\,2j\,+\,j^2\right)\,\left(1\,-\,j\right)\,=\,-2j\left(1\,-\,j\right)\,=\,-2j\,+\,2j^2\,=\,-2\,-\,2j$$

g) $$\left(1\,-\,j\right)^{\frac{1}{2}}\,=\,\left(\sqrt{2}\,\angle\,-45^{\circ}\right)^{\frac{1}{2}}\,=\,(2)^{\frac{1}{4}}\,\angle\,-22.5^{\circ}$$

Look good?

 

[olgranpappy]

g) $$\left(1\,-\,j\right)^{\frac{1}{2}}\,=\,\left(\sqrt{2}\,\angle\,-45^{\circ}\right)^{\frac{1}{2}}\,=\,(2)^{\frac{1}{4}}\,\angle\,-22.5^{\circ}$$

Look good?

well... is that in rectangular form?

 

[Gib Z]

well... is that in rectangular form?

It doesn't seem to be of any form that I know of

VinnyCee- You do know that $$\left( \exp (i \theta) \right)^{1/2} = \exp (i \cdot \frac{\theta}{2} )$$ ? That is easy to see if you take the exponential being used here as the exponential function extended to the complex numbers, though if you are merely using exp(ix) as a formal abbreviation for cos x + i sin x, then you must use De Moirves theorem to see it ( well, actually, the generalization of it to non-integer exponents).

 

[olgranpappy]

It doesn't seem to be of any form that I know of

I figured he meant that he was specifying the magnitude of the number and the phase. which would be correct, but he has to expand the exponential in terms of cos and sin to get the "rectangular form".

 

[VinnyCee]

OK - If I convert (g)...

$$2^{\frac{1}{4}}\,\angle\,-22.5^{\circ}\,=\,2^{\frac{1}{4}}\,cos\left(-22.5^{\circ}\right)\,+\,j\,2^{\frac{1}{4}}\,sin\left(-22.5^{\circ}\right)\,\approx\,1.099\,-\,0.455\,j$$

Now, do these seem right, (a) through (g)? If not, can you show me the errors?

 

[olgranpappy]

OK - If I convert (g)...

$$2^{\frac{1}{4}}\,\angle\,-22.5^{\circ}\,=\,2^{\frac{1}{4}}\,cos\left(-22.5^{\circ}\right)\,+\,j\,2^{\frac{1}{4}}\,sin\left(-22.5^{\circ}\right)\,\approx\,1.099\,-\,0.455\,j$$

Now, [does this] seem right...

looks fine.