# Complex Numbers - a review problem (a - g)

1. Jan 13, 2008

### VinnyCee

1. The problem statement, all variables and given/known data

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

2. Relevant equations

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

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

where $z^*$ is the complex conjugate

stuff like that...

3. 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...

2. Jan 13, 2008

### olgranpappy

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

3. Jan 13, 2008

### VinnyCee

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

4. Jan 13, 2008

### HallsofIvy

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

5. Jan 13, 2008

### olgranpappy

I dont 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.

6. Jan 14, 2008

### 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?

7. Jan 14, 2008

### olgranpappy

well... is that in rectangular form?

8. Jan 14, 2008

### Gib Z

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).

9. Jan 15, 2008

### olgranpappy

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".

10. Jan 15, 2008

### 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?

11. Jan 16, 2008

looks fine.