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August 20, 2014

August 20, 2014

Posted by **Kerry-Ann** on Friday, September 13, 2013 at 6:25pm.

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[1-itan(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 7:17pm1. Let a = x + yi for reals x and y

Thus solve the simultaneous equations:

x^2-y^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 = -2-5i

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 = 3^2-2^2,

(3-2i)^2 = 9-12i-4 = 5-12i

(-3+2i)^2 = 5-12i

Or, you can do it by equating real/imaginary parts:

(a+bi)^2 = (a^2-b^2) + 2abi

so

a^2-b^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+1-2i)

so,

z = -2-5i or -1-2i or -1+2i

- PURE MATHEMATICS -
**Kerry-Ann**, Friday, September 13, 2013 at 7:37pmfor 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 7:45pmSteve, #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 7:47pmHmmm. Makes sense. Thanks for checking. I forgot we were on the unit circle.

- PURE MATHEMATICS -
**Kerry-Ann**, Friday, September 13, 2013 at 8:00pmso number 8 is has trigonometry identities in it?

- PURE MATHEMATICS -
**Kerry-Ann**, Friday, September 13, 2013 at 8:07pmthanks 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 -
**Kerry-Ann**, Friday, September 13, 2013 at 9:17pmExplain 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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