precalculus honors

Question

Find all complex numbers z such that z^2 = 2i.

Answer 1

... or using de Moivre's Theorem:  w_k =  z^1/n = (rcisθ)^1/n = r^1/n (cis(θ+360k)/n), where k = 0 and 1, and n = 2,

z = 0 + 2i, so r = 2, and θ = 90°.

 

we get w₀ = √2 (cis(90/2)), and w₁ = √2 (cis(90 + 360)/2);

 

So, w₀ = √2 cos 45 + √2 i sin 45 = 1 + i

 

and w₁ = √2 cos 225 + √2 i sin 225 = -1 - i

 

Answer 2

z²=2i

-z²=-2i

-z²=-2i+1-1

-z²=1-2i-1

-z²=1²-2i-1²

-z²=(1-i)²

√-z²=√(1-i)²

√-1√z²=√(1-i)²

(i)(z)=+(1-i)

z=(1-i)/i

z=-(1-i)/i

z=(1-i)(i)/(i)(i)

z=i-i²/i²

z=i+1/(-1)

z=-i-1

z=-1-i

z=-(1-i)(i)/(i)(i)

z=-i+i²/i²

z=-i-1/-1

z=i+1

z=1+i

Answer 3

Let z = a+bi where z² = 2i

 

(a+bi)² = 0+2i

 

a² + 2abi - b² = 0 + 2i

 

(a² - b²) + 2abi = 0 + 2i

 

Equating the real parts and imaginary parts, we arrive at the system of equations:

 

a² - b² = 0

2ab = 2

 

Rewriting, we have a² = b²

                                   ab = 1   So, a = 1/b

 

Replace a by 1/b in the equation a² = b²:     1/b² = b²

 

Therefore, b⁴ = 1     So, b⁴ - 1 = 0

                                         (b² -1)(b² +1) = 0

 

 b² - 1 = 0 or b² + 1 = 0

 

b is a real number and b² + 1 = 0 has no real solutions, so we are left with       b² - 1 = 0.

 

Thus, b = 1 or -1

 

If b = 1, then a = 1/b = 1

If b = -1, then a = 1/b = -1

 

If z² = 2i, then z = 1 + i or -1 - i 

 

NOTE:  an alternate method for solving this problem is to use DeMoivre's Theorem.