Asked by Kerry-Ann
1. The complex number 'a' is such that a^2=5-12i
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
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
Answers
Answered by
Graham
1. 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.
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.
Answered by
Steve
#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
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
Answered by
Kerry-Ann
for the first one I got a to be -0.67 and 1.5 when b is 9 and -4, respectively
Answered by
Graham
Steve, #8 is true. Your example is wrong.
Note: if θ=(π/4) then z = (1/√2) + (i/√2)
Note: if θ=(π/4) then z = (1/√2) + (i/√2)
Answered by
Steve
Hmmm. Makes sense. Thanks for checking. I forgot we were on the unit circle.
Answered by
Kerry-Ann
so number 8 is has trigonometry identities in it?
Answered by
Kerry-Ann
thanks very much for helping with 1 and 9
number 8 is still giving me some trouble though
number 8 is still giving me some trouble though
Answered by
Graham
#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.
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.
Answered by
Kerry-Ann
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?
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.