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!
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 $\displaystyle -\pi < Arg(w) < \pi$.)
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
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