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Find all complex numbers $z$ such that $|z|^2-2\bar z+iz=2i.$

Apr 16, 2018

#1
+111329
+3

$$|z|^2-2\bar z+iz=2i.$$

z*z(bar) - 2(a - bi) + i (a + bi)  =  2i

(a + bI) ( a - bi) - 2a + 2bi + ai + bi^2  =  2i

a^2+ b^2 - 2a + 2bi + ai + bi^2  = 2i

a^2 + b^2 - 2a + 2bi + ai - b  = 2i

(a^2 + b^2 - 2a - b) + (2b + a)i  =   0 +  2i     equate coefficients

Thus

a^2 + b^2 - 2a - b  =  0   (1)

2b + a  = 2  ⇒  a  = 2 - 2b  ⇒  a = 2 (1 - b)   (2)

Sub (2)  into (1)

[4 (1 - b)^2] + b^2 - 2 [ 2 (1 - b) ] - b  = 0

4 (b^2 - 2b + 1) + b^2 - 4 + 4b  - b  = 0

4b^2 - 8b + 4 + b^2 - 4 + 4b - b  = 0

5b^2 - 5b = 0

b^2 - b  = 0

b(b - 1)  = 0      set each factor to 0 and solve for b

b  =  0     or  b  = 1

So when b = 0, a  =  2 (1 - 0)  = 2

And when b  = 1,   a  = 2 ( 1 - 1)  = 0

So  the solutions are

z = 2 + 0i ,   z = 0 +1i

Apr 16, 2018
edited by CPhill  Apr 16, 2018
#2
+24983
+2

Find all complex numbers $z$ such that

$$|z|^2-2\bar z+iz=2i.$$

|z|^2-2\bar z+iz=2i.

$$\begin{array}{|rcll|} \hline |z|^2-2\bar z+iz &=& 2i \quad & | \quad |z|^2=z\bar z \\ z\bar z-2\bar z+iz &=& 2i \quad & | \quad -iz \\ z\bar z-2\bar z &=& 2i -iz \\ \bar z(z-2) &=& i(2 - z) \\ \bar z(z-2) &=& -i(z - 2) \quad & | \quad z-2 \ne 0 \\ \bar z &=& -i\left(\dfrac{z - 2}{z-2}\right) \\ \bar z &=& -i \quad & | \quad \bar z = a-bi \\\\ a-bi &=& 0-i \quad & | \quad \text{compare both sides} \\ && \boxed{a=0\qquad b = 1 \qquad \text{or} \qquad z= 0 + 1i} \\\\ \bar z(z-2) &=& -i(z - 2) \quad & | \quad z-2 = 0 \\ z-2 &=& 0 \quad & | \quad + 2 \\ z &=& 2 \quad & | \quad z = a+bi \\\\ a+bi &=& 2+0i \quad & | \quad \text{compare both sides} \\ && \boxed{a=2\qquad b = 0 \qquad \text{or} \qquad z= 2 + 0i} \\ \hline \end{array}$$

Apr 16, 2018