quick qu --> complex numbers (1 Viewer)

[sikeveo]

Having a little trouble on these. I keep getting totally wrong answers.

 

[Slidey]

a)
z^2=r^2.e^2i@
Times by (1-i) which is sqrt2.e^(-i.pi/4)
To get:
(1-i).z^2=sqrt2.r^2.e^(i[2@-pi/4])
Mod of both sides:
|(1-i).z^2|=sqrt.r^2
Why? Because be definition, z=re^(i@), |z|=r.

b) From above,
(1-i).z^2=sqrt2.r^2.e^(i[2@-pi/4])
Again by definition:
Arg((1-i).z^2)=2@-pi/4

Similarly, for c and d:
(1+i.sqrt3)/z=2.e^(i.pi/3)/[r.e^(i@)]=2e^(i[pi/3-@])/r
Now take the mod and arg of both sides.

 

[KFunk]

It might be helpful to start with some of the mod/arg properties e.g. |zw| = |z|.|w| and arg(zw) = arg(z) + arg(w) so for the first one you have:

a) |(1-i)z²| = |(1-i)|.|z²| = (√2)r²

^since |(1 - i)| is √2 and |z| = r --> |z²| = r² (note that e^iθ = cisθ )

b) arg[(1-i)z²] = arg(1-i) + arg(z²) = -π/4 + 2θ

c) |1 + i√3|/|z| ... etc

d) arg(1 + i√3) + arg(z) ... etc

 

[MAICHI]

Those questions look like it came out of math 1131.

Remember one thing though, keep the your argument in the range (-pi, pi], never assume they are.

 

[sikeveo]

thanks maichi, you're correct.

 

[Riviet]

Isn't this stuff uni maths?