Complex Number Question for 2014ers (1 Viewer)

[hit patel]

i) If tan theta = 1/3, show that tan 4theta = 24/7 and hence find the fourth root of 7+24i of smallest argument in polar form where a and b are real.
ii) Derive an expression for tan 4theta and solve tan4theta= 1/2 +sqrt(3)i/2

 

[Carrotsticks]

i) If tan theta = 1/3, show that tan 4theta = 24/7 and hence find the fourth root of 7+24i of smallest argument in polar form where a and b are real.
ii) Derive an expression for tan 4theta and solve tan4theta= 1/2 +sqrt(3)i/2

Solving trig equations with complex values is not in the syllabus.

 

[anomalousdecay]

Solving trig equations with complex values is not in the syllabus.

^this. Only part (i) is in the syllabus, but would take very long if in HSC and would be a 3 marker in the last question at most. You should focus on the foundations at the moment for the students only starting the syllabus OP.

 

[hit patel]

How come ?! Isnt it in the syllabus? oh,.,... . ok right. yep then they can do the second part.

 

[emilios]

After proving the first part of i), tan(4\theta) = 24/7 (pretty straightfoward), this is my working: (theta has been changed to x for brevity)

let (7+24i)^1/4 = z
convert to mod-arg form; (25cis(24/7))^1/4 = z
z= sqrt5*cis(tan4x))^1/4
z= sqrt5cis(tanx))

Now, is it right to assume tanx = 1/3? If so then:
z = sqrt5cis(1/3). And that is the smallest argument (we could add 2kpi, but this is the smallest).

 

[hit patel]

Ok first off all it is not obvious and you get a great deal of 3-4 marks to prove the first statement. But excellent work on the second part. Polar form would have been better but its ok.