Posted by KerryAnn on Friday, September 13, 2013 at 6:25pm.
1. The complex number 'a' is such that a^2=512i
use an algebraic method to find the two possible values of 'a'
8. The complex number z is given by z=cosTheta + isinTheta where 1/2pi< or equal to Theta less than or equal to 1/2pi.
a. show that 1+z=2cos(1/2Theta) and arg(1+z)=1/2Theta.
b. hence, show that 1/(1+z)=1/2[1itan(1/2Theta)]
c. describe the locus of the point representing z and the locus of the point representing 1/(1+z) in an argand diagram as Theta varies from pi/2 to pi/2
9. A) find the exact value of all the roots of the equation (z+2+5i)(z^2+2z+5)=0

PURE MATHEMATICS  Graham, Friday, September 13, 2013 at 7:17pm
1. Let a = x + yi for reals x and y
Thus solve the simultaneous equations:
x^2y^2 = 5, and 2xy = 12
8. 1+z = 1+cos(θ) + i sin(θ)
Hint: Substitute θ=2ω then simplify
Use: sin(2ω) = 2 sin(ω) cos(ω)
And: cos(2ω) = 2 cos^2(ω)  1
9. One root: z = 25i
The other roots will be those of:
z^2 + 2z + 5 = 0
Hint: Use either complete the square, or the quadratic formula.

PURE MATHEMATICS  Steve, Friday, September 13, 2013 at 7:33pm
#1
since 5 = 3^22^2,
(32i)^2 = 912i4 = 512i
(3+2i)^2 = 512i
Or, you can do it by equating real/imaginary parts:
(a+bi)^2 = (a^2b^2) + 2abi
so
a^2b^2 = 5
2ab = 12
a^2  (6/a)^2 = 5
and solve for a
#8 I don't think this is true.
If z = 1+i, θ=pi/4
1+z = 2+i
1+z=√5 and arg(z+1) = arctan(1/2) which is not pi/8
#9
z^2+2z+5 = (z+1)^2 + 2^2 = (z+1+2i)(z+12i)
so,
z = 25i or 12i or 1+2i

PURE MATHEMATICS  KerryAnn, Friday, September 13, 2013 at 7:37pm
for the first one I got a to be 0.67 and 1.5 when b is 9 and 4, respectively

PURE MATHEMATICS  Graham, Friday, September 13, 2013 at 7:45pm
Steve, #8 is true. Your example is wrong.
Note: if θ=(π/4) then z = (1/√2) + (i/√2)

PURE MATHEMATICS  Steve, Friday, September 13, 2013 at 7:47pm
Hmmm. Makes sense. Thanks for checking. I forgot we were on the unit circle.

PURE MATHEMATICS  KerryAnn, Friday, September 13, 2013 at 8:00pm
so number 8 is has trigonometry identities in it?

PURE MATHEMATICS  KerryAnn, Friday, September 13, 2013 at 8:07pm
thanks very much for helping with 1 and 9
number 8 is still giving me some trouble though

PURE MATHEMATICS  Graham, Friday, September 13, 2013 at 8:41pm
#8. 1+z = 1+cos(θ) + i sin(θ)
Substitute θ=2ω then simplify
Use: sin(2ω) = 2 sin(ω) cos(ω)
And: cos(2ω) = 2 cos^2(ω)  1
So:
1+z = 1+cos(2ω) + i sin(2ω)
1+z = 2 cos^2(ω) + 2i sin(ω) cos(ω)
1+z = 2 cos(ω) (cos(ω) + i sin(ω))
1+z = 2 cos(θ/2)(cos(θ/2) + i sin(θ/2))
Thus for π/2 ≤ θ ≤ π/2:
1+z = 2 cos(θ/2)
arg(1+z) = θ/2
Then for (b) use:
1/(cos(ω)+i sin(ω)) = cos(ω)i sin(ω)
To find 1/(1+z)
Then for (c) plot z and 1/(1+z) on the complex plane and describe.

PURE MATHEMATICS  KerryAnn, Friday, September 13, 2013 at 9:17pm
Explain to me just this one line 1+z = 2 cos(θ/2)(cos(θ/2) + i sin(θ/2))what happened to i sin(θ/2)) in the other two lines which follow?
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