patel complex help (1 Viewer)

[CriminalCrab]

I need help with questions 10 and 11 in exercise 4K from the patel book.
Here are the questions:

10. find the modulus and argument of each of the complex numbers z and w.
z= (1+i)/(1-i) and w=(root2)/(1-i)
Plot the points representing z, w and z+w on argand diagram. Deduce, from the diagram that tan(3pi/8)= root2 + 1

11. The points P and Q are represented by the complex numbers z=1-3i and w=-3+4i respectively. Find the point R on the real axis such that PRQ is a right angled triangle.
answer: (3,0) or (-5,0)

thanks in advance

 

[asianese]

z=(1+i)/(1-i) * (1+i)/(1+i) realising denom = i (after simplification)
z=cis(pi/2)

w=root2/1-i * (1+i)/(1+i) realising denom = 1/root2 + 1/root2 * i

z+w = i+ 1/root2 + 1/root2 i
=root2 +1 after simpliciation

tantheta = (1+1/root2)/(1/root2) = root2+1

theta = 67.5 = 3pi/8 after inverse tanning

11.

just use simple locus.

if they are at 90 degs, then the gradients are negative inverses.

Mqr * Mpr = -1
4/-3-x * 3/x-1 = -1
12=-((-3-x)(x-1))
...
...
x^2 + 2x -15 =0
x=3 or -5

 

[SpiralFlex]

Here is the corresponding pretty diagram to go with it. Rushed, but still pretty.

 

[SpiralFlex]

Spiralflex, can you tell me how you did that, cause i'd be much better helper then haha.
thanks, and gl with hsc (are u a furture stateranker in ext 2?)

I used MSpaint.

 

[AAEldar]

yea, how did u get it like into the post haha, soz for being noob

Upload the picture to a site like imgur.com or imageshack.us and copy the link with the

 

[CriminalCrab]

I get it now.
Thanks people =)

 

[bleakarcher]

I have just "realised" the denominator in both cases, sort of like rationalizing them
Determining the modulus and argument is crucial to solving this mystery:





now when you plot z and w on an angrad diagram (i don't yet know a tool for how to show you this plot sorry, if someone could point me to it), the arg(z+w) = pi/2 + pi/4 = 3pi/8, so you just use simple trig, where tan(3pi/8) = the vertical component of the z+w vector(which requires additional trig for discovery) over the horizontal component of the w vector (more trig required) this is basically tan(x) = oppisite/adjacent side








Since PRQ is right angled triangle

Solving this yields x =3,-5 soz, cbb expanding, if you can't expand ask, i'll do it.

Hope that helps, if there is something not understood ask, and rep please, took me long

Lolwut...?