# Complex Numbers - Help please.

• Aug 13th 2006, 05:03 AM
c00ky
Hello there,
I'm really struggling with the following problem and would be VERY grateful if somebody could help me out, i've done some work but i want to check if i'm on the right track.

Consider the following Complex variables

(i) 2 + j5
(ii) - 4 + j10
(iii) - 6 - j3

Questions:

1) Multiply (i) and (ii) above by first converting to Polar form and express the result in Polar form.

2) Divide (ii) by (iii) above by first converting to Polar form and express the result in Polar form.
• Aug 13th 2006, 05:18 AM
topsquark
Quote:

Originally Posted by c00ky
Hello there,
I'm really struggling with the following problem and would be VERY grateful if somebody could help me out, i've done some work but i want to check if i'm on the right track.

Consider the following Complex variables

(i) 2 + j5
(ii) - 4 + j10
(iii) - 6 - j3

Questions:

1) Multiply (i) and (ii) above by first converting to Polar form and express the result in Polar form.

2) Divide (ii) by (iii) above by first converting to Polar form and express the result in Polar form.

As a start, convert all three to polar form:

i) This has the coordinates (2,5) in the Argand plane. Thus the "length" of this number is $\sqrt{2^2+5^2}=\sqrt{29} \approx 5.385$. The "angle" will be $tan^{-1} \left ( \frac{5}{2} \right ) \approx 1.190 \, rad$, so the polar form of 2 + j5 is approximately $5.385e^{1.190j}$.

Similarly
ii) -4 + j10 is approximately $10.770e^{1.951j}$.

iii) -6 - j3 is approximatetly $6.7082e^{3.605j}$.

Multiply i) and ii). We are multiplying: $ae^{bj} \cdot ce^{dj} = (ac)e^{(b+d)j}$, by the rules of exponents.

Division is similar: $ae^{bj} \div ce^{dj} = \left ( \frac{a}{c} \right ) e^{(b-d)j}$.

-Dan
• Aug 13th 2006, 05:55 AM
c00ky
I'm working in degrees not radians.

Just a quick question because i'm totally lost here...

Consider the following complex number:

3-j6

in it's polar form would this be

6.7 /296.57

???
• Aug 13th 2006, 06:13 AM
topsquark
Quote:

Originally Posted by c00ky
I'm working in degrees not radians.

Just a quick question because i'm totally lost here...

Consider the following complex number:

3-j6

in it's polar form would this be

6.7 /296.57

???

You got it.

In any event, the method I showed you works for either degrees or radians. Though I believe that the unit "radian" is standard for the form I showed you, you can obviously just convert from radians to degrees.

-Dan
• Aug 13th 2006, 07:05 AM
Soroban
Hello, c00ky!

I'll run through #1 . . .

Quote:

Consider the following complex numbers:

$(a)\;2+5i\qquad (b)\;-4+10i\qquad(c)\;-6-3i$

1) Multiply (a) and (b) above by first converting to Polar form
. . .and express the result in Polar form.

We have:
. . $(a)\;2+5i \;= \;\sqrt{29}(\cos\alpha + i\sin\alpha)$ . . . where $\alpha = \arctan\left(\frac{5}{2}\right)$
. . $(b)\;\text{-}4+10i\;=\;2\sqrt{29}(\cos\beta + i\sin\beta)$ . . . where $\beta = \arctan\left(\text{-}\frac{5}{2}\right)$

We're expected to know the product of two complex numbers in polar form:
. . . . $\begin{array}{cc}\text{If }z_1 \:=\:r_1(\cos\alpha + i\sin\alpha)\text{ and }z_2\:=\:r_2(\cos\beta + i\sin\beta) \\
\text{then: }z_1z_2\;=\;r_2r_2\left[\cos(\alpha + \beta) + i\sin(\alpha + \beta)\right] } \end{array}$

We have: . $\left[\sqrt{29}(\cos\alpha + i\sin\alpha)\right]\,\left[2\sqrt{29}(\cos\beta + i\sin\beta)\right]$

. . . . . . $= \;58\left[\cos(\alpha + \beta) + i\sin(\alpha + \beta)\right]$

Now $\alpha = \arctan\left(\frac{5}{2}\right)$ is in quadrant 1 . . . And $\beta = \arctan\left(\text{-}\frac{5}{2}\right)$ is in quadrant 2.
. . If you make a sketch, you will see that: . $\alpha + \beta \,=\,180^o$

Therefore, the product is: . $\boxed{58(\cos180^o + i\sin180^o)}$