# Complex Number's Assignment (From my signals class)

1. Sep 4, 2006

### Naeem

Q1. If j = square root of - 1 what is square root of j ?

what I did was : Plug in the value of j in square root of j and came up with -1 to the power of 0.25.

Is this right. Looks like this wrong.

Q2. Given a complex number w = x + jy, the complex congugate of w is defined in rectangular coordinates as w* = x-jy. Use this fact to derive complex congugation in polar form.

What I did was : multiply both w * w* and came up with x ^ 2 + y ^2

and I know euler's formula e ^(jtheta) = cos (theta) + i sin (theta)

Is this right, probably not, can someone guide me here as well.

Q3. By hand sketch the following against independent variable t:

(a) x2(t) = Im (3 - e(1-j2pi)t)

There is another two also in these parts, One with the real part given and another one with x3(t) = 3 - Im(e(1-j2pi)t)

How do I do these problems ? Please any one help me. The book is worthless. It just talks about basics on complex numbers, congugates , polar forms etc.

But this HW has been a pain belive me....

2. Sep 4, 2006

### rbj

i think your prof wants to know what the real part and imaginary part of $\sqrt{j}$ is. (maybe magnitude and angle, but that would be too easy.) remember, the square root of anything has two solutions.

you are being asked to express $w$ in polar form and $w^*$ in terms of $w$ in polar form. i think that is the case.

what do you know about the exponential? what happens when the exponent is the sum of two terms? what do you do with it?

try using $\LaTeX$ to express your math here on this forum. we ain't USENET here where you need to rely on "ASCII math".

Last edited: Sep 4, 2006
3. Sep 12, 2006

### ravenprp

$\sum_{k=1}^n k^2 = \frac{1}{2} n (n+1).$

4. Sep 13, 2006

### Corneo

For Q2: Say your given $w = x + jy$. Arbitrarily draw this on the x-y plane. Now draw $w^*$. Express w in polar form. What do you notice about w* in polar form?

5. Sep 13, 2006

### Corneo

For Q1: One way to solve this is express j in polar form. You can get this from Euler's identiy and see that $e^{j \pi /2} = \cos (\pi/2)+ j \sin (\pi/2) = j$. Then you have

$$\sqrt{j} = \sqrt{e^{j \pi /2}}$$