Complex Numbers (2 Viewers)

[Kujah]

Need some help:

1. By solving x^2 - 2(1+i)z + 8i= 0, show that (1+root3) +i(1-root3) and (1-root3)+i(1+root3) are the roots.

I've tried using the quadratic formula again and again, but it's gotten me nowhere.

2. By expressing cos4Θ and sin4Θ in terms of powers of cosΘ and sinΘ, show that:

tan4Θ = 4tanΘ-4tan3Θ/1-6tan²Θ +tan4Θ

I've expanded it out, and equated real and imaginary parts. Then what do I do?

Thanks for any help ^<o></o>

 

[Trebla]

For 1), you could use sum and product of roots but since you have to 'solve' it, you use the quadratic formula which leads to you having to find the square root of a complex number.
From the quadratic formula x = [2(1 + i) ± √(- 24i)]/2
=> x = (1 + i) ± √(- 6i)
=> x = (1 + i) ± i √(6i) *
We must find another expression for √(6i).
Let: √(6i) = (a + ib)
6i = (a + ib)²
6i = a² - b² + 2abi
equate real and imaginary parts
a² = b²
ab = 3
=> a = b = ±√3
.: √(6i) = ±(√3 + i√3) = ± √3 (1 + i)

From * we get
x = (1 + i) ± i √3 (1 + i)
=> x = (1 + i) ± √3 (- 1 + i)
Take positive case: x = 1 + i - √3 + i√3 = (1 - √3) + i (1 + √3)
Take negative case: x = 1 + i + √3 - i√3 = (1 + √3) + i (1 - √3)

 

[BBB237]

OMG. Your all crazy. How do you understand all of that?

 

[velox]

OMG. Your all crazy. How do you understand all of that?

It's a pretty stock question....

 

[BBB237]

Your point?

 

[Kujah]

Thanks Trebla.

I've done the second question