Let z = x + i y z = x + iy z = x + i y x , y ∈ R x, y \in \mathbb{R} x , y ∈ R
Given ∣ z ∣ = 2 |z| = \sqrt{2} ∣ z ∣ = 2 ∣ z ∣ 2 = x 2 + y 2 = 2. |z|^2 = x^2 + y^2 = 2. ∣ z ∣ 2 = x 2 + y 2 = 2.
Also, ∣ z 2 − i ∣ = 1 |z^2 - i| = 1 ∣ z 2 − i ∣ = 1 z 2 z^2 z 2 z 2 = ( x + i y ) 2 = x 2 − y 2 + 2 i x y . z^2 = (x + iy)^2 = x^2 - y^2 + 2ixy. z 2 = ( x + i y ) 2 = x 2 − y 2 + 2 i x y . z 2 − i = ( x 2 − y 2 ) + 2 i x y − i = ( x 2 − y 2 ) + i ( 2 x y − 1 ) . z^2 - i = (x^2 - y^2) + 2ixy - i = (x^2 - y^2) + i(2xy - 1). z 2 − i = ( x 2 − y 2 ) + 2 i x y − i = ( x 2 − y 2 ) + i ( 2 x y − 1 ) . ∣ z 2 − i ∣ 2 = ( x 2 − y 2 ) 2 + ( 2 x y − 1 ) 2 = 1. |z^2 - i|^2 = (x^2 - y^2)^2 + (2xy - 1)^2 = 1. ∣ z 2 − i ∣ 2 = ( x 2 − y 2 ) 2 + ( 2 x y − 1 ) 2 = 1.
Now, we have the system:x 2 + y 2 = 2 ( x 2 − y 2 ) 2 + ( 2 x y − 1 ) 2 = 1 \begin{align*}
x^2 + y^2 &= 2 \\
(x^2 - y^2)^2 + (2xy - 1)^2 &= 1
\end{align*} x 2 + y 2 ( x 2 − y 2 ) 2 + ( 2 x y − 1 ) 2 = 2 = 1
Let us use polar form: z = r e i θ z = r e^{i\theta} z = r e i θ r = 2 r = \sqrt{2} r = 2 z = 2 e i θ z = \sqrt{2} e^{i\theta} z = 2 e i θ z 2 = 2 e i 2 θ z^2 = 2 e^{i2\theta} z 2 = 2 e i 2 θ z 2 − i = 2 e i 2 θ − i . z^2 - i = 2 e^{i2\theta} - i. z 2 − i = 2 e i 2 θ − i .
Write 2 e i 2 θ = 2 ( cos 2 θ + i sin 2 θ ) 2 e^{i2\theta} = 2(\cos 2\theta + i \sin 2\theta) 2 e i 2 θ = 2 ( cos 2 θ + i sin 2 θ ) z 2 − i = 2 cos 2 θ + i ( 2 sin 2 θ − 1 ) . z^2 - i = 2\cos 2\theta + i(2\sin 2\theta - 1). z 2 − i = 2 cos 2 θ + i ( 2 sin 2 θ − 1 ) .
Therefore,∣ z 2 − i ∣ 2 = ( 2 cos 2 θ ) 2 + ( 2 sin 2 θ − 1 ) 2 = 1. |z^2 - i|^2 = (2\cos 2\theta)^2 + (2\sin 2\theta - 1)^2 = 1. ∣ z 2 − i ∣ 2 = ( 2 cos 2 θ ) 2 + ( 2 sin 2 θ − 1 ) 2 = 1. 4 cos 2 2 θ + ( 4 sin 2 2 θ − 4 sin 2 θ + 1 ) = 1 4 cos 2 2 θ + 4 sin 2 2 θ − 4 sin 2 θ + 1 = 1 4 ( cos 2 2 θ + sin 2 2 θ ) − 4 sin 2 θ + 1 = 1 4 ( 1 ) − 4 sin 2 θ + 1 = 1 4 − 4 sin 2 θ + 1 = 1 4 − 4 sin 2 θ = 0 1 − sin 2 θ = 0 sin 2 θ = 1 \begin{align*}
4\cos^2 2\theta + (4\sin^2 2\theta - 4\sin 2\theta + 1) &= 1 \\
4\cos^2 2\theta + 4\sin^2 2\theta - 4\sin 2\theta + 1 &= 1 \\
4(\cos^2 2\theta + \sin^2 2\theta) - 4\sin 2\theta + 1 &= 1 \\
4(1) - 4\sin 2\theta + 1 &= 1 \\
4 - 4\sin 2\theta + 1 = 1 \\
4 - 4\sin 2\theta = 0 \\
1 - \sin 2\theta = 0 \\
\sin 2\theta = 1
\end{align*} 4 cos 2 2 θ + ( 4 sin 2 2 θ − 4 sin 2 θ + 1 ) 4 cos 2 2 θ + 4 sin 2 2 θ − 4 sin 2 θ + 1 4 ( cos 2 2 θ + sin 2 2 θ ) − 4 sin 2 θ + 1 4 ( 1 ) − 4 sin 2 θ + 1 4 − 4 sin 2 θ + 1 = 1 4 − 4 sin 2 θ = 0 1 − sin 2 θ = 0 sin 2 θ = 1 = 1 = 1 = 1 = 1
So 2 θ = π 2 + 2 π k 2\theta = \frac{\pi}{2} + 2\pi k 2 θ = 2 π + 2 π k k ∈ Z k \in \mathbb{Z} k ∈ Z θ = π 4 + π k \theta = \frac{\pi}{4} + \pi k θ = 4 π + π k
Therefore, the solutions arez = 2 e i ( π 4 + π k ) z = \sqrt{2} e^{i\left(\frac{\pi}{4} + \pi k\right)} z = 2 e i ( 4 π + π k ) k ∈ Z k \in \mathbb{Z} k ∈ Z
Explicitly, for k = 0 k = 0 k = 0 z 1 = 2 e i π 4 = 2 ( cos π 4 + i sin π 4 ) = 2 ( 2 2 + i 2 2 ) = 1 + i . z_1 = \sqrt{2} e^{i\frac{\pi}{4}} = \sqrt{2}\left(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}\right) = \sqrt{2}\left(\frac{\sqrt{2}}{2} + i \frac{\sqrt{2}}{2}\right) = 1 + i. z 1 = 2 e i 4 π = 2 ( cos 4 π + i sin 4 π ) = 2 ( 2 2 + i 2 2 ) = 1 + i . k = 1 k = 1 k = 1 z 2 = 2 e i ( 5 π 4 ) = 2 ( cos 5 π 4 + i sin 5 π 4 ) = 2 ( − 2 2 − i 2 2 ) = − 1 − i . z_2 = \sqrt{2} e^{i\left(\frac{5\pi}{4}\right)} = \sqrt{2}\left(\cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4}\right) = \sqrt{2}\left(-\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}\right) = -1 - i. z 2 = 2 e i ( 4 5 π ) = 2 ( cos 4 5 π + i sin 4 5 π ) = 2 ( − 2 2 − i 2 2 ) = − 1 − i .
Thus, the solutions are z = 1 + i z = 1 + i z = 1 + i z = − 1 − i z = -1 - i z = − 1 − i