hi...

appreciate if anyone can help me with this complex no. question....especially bi)....need detailed explanation!

Ans as follows:

(b) (i) 2.21

(ii) -2 + I , -2 -i

Thank You!

Printable View

- Oct 22nd 2006, 01:58 AMshaojie2kComplex No.
hi...

appreciate if anyone can help me with this complex no. question....especially bi)....need detailed explanation!

Ans as follows:

(b) (i) 2.21

(ii) -2 + I , -2 -i

Thank You! - Oct 22nd 2006, 04:05 AMtopsquark
- Oct 22nd 2006, 04:20 AMtopsquark
a)

$\displaystyle \frac{w^2+4}{4-w^2} = i$

$\displaystyle w^2 + 4 = 4i - iw^2$

$\displaystyle (1+i)w^2 = -4 + 4i$

$\displaystyle w^2 = 4 \frac{-1+i}{1+i}$

The simplest way to proceed from here would be to transform the RHS to the form $\displaystyle re^{i \theta}$.

$\displaystyle \frac{-1+i}{1+i} = \frac{-1+i}{1+i} \cdot \frac{1-i}{1-i} = \frac{-(1-i)^2}{1+1} = -\frac{1-2i-1}{2} = i = e^{\pi \, i/2}$

So

$\displaystyle w^2 = e^{\pi \, i/2}$

$\displaystyle w = \pm e^{\pi \, i/4}$

So one solution for w is $\displaystyle e^{\pi \, i/4}$ and the other is $\displaystyle -e^{\pi \, i/4} = e^{\pi \, i} \cdot e^{\pi \, i/4} = e^{5 \pi \, i/4} = e^{-2 \pi \, i} \cdot e^{5 \pi \, i/4} = e^{-3 \pi \, i/4}$. (The last trick with the $\displaystyle e^{-2 \pi \, i}$ is simply to force $\displaystyle -\pi < Arg(w) < \pi$.) - Oct 22nd 2006, 07:51 AMtopsquark
i)

I'm having a problem with this. My approach is to write $\displaystyle z + i = re^{i \theta} + i = Re^{i \Theta}$

I got that $\displaystyle \Theta = atn \left ( \frac{1+sin \theta}{cos \theta} \right )$

The problem is that this function has no minimum:

$\displaystyle \frac{d \Theta}{d \theta} = \frac{1}{\left ( \frac{1+sin \theta}{cos \theta} \right )^2 + 1} \cdot \left [ 1 + \frac{sin^2 \theta + sin \theta}{cos^2 \theta} \right ] $ = $\displaystyle \frac{1 + sin \theta}{2(sin \theta + 1)} = \frac{1}{2}$

which can never be zero.

Now, looking at the graph of the function $\displaystyle \Theta( \theta )$ you might think that there is a local minimum at $\displaystyle \theta = \pi /2$, but the function does not exist there.

(I'm going to attempt to upload the graph. Cross your fingers!)

(Yay! It worked! :) )

-Dan - Oct 22nd 2006, 07:58 AMtopsquark
ii) Again, I'm having a problem.

Write $\displaystyle z = re^{i \, \theta} = r \, cos \theta + ir \, sin \theta$

Then $\displaystyle z - z^* = 2ir \, sin \theta$

$\displaystyle |z - z^*| = 2r |sin \theta |$

This has no maximum since we may choose r to be as large as we wish.

-Dan