Complex numbers again

Algebra Level 4

If z z be a complex number satisfying ∣ z ∣ 2 + 2 ( z + z ‾ ) + 3 i ( z − z ‾ ) + 4 = 0 { |z| }^{ 2 }+2(z+\overline { z } )+3i(z-\overline { z } )+4=0 , then complex number z + 3 + 2 i z+3+2i will lie on a circle with centre a + b i a+bi and radius c c units.

Find ∣ a + b + c ∣ |a+b+c| .


The answer is 9.

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Tanishq Varshney by Tanishq Varshney , 23, India

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2 solutions

Tanishq Varshney
May 23, 2015

∣ z ∣ 2 + ( 2 + 3 i ) z + ( 2 − 3 i ) z ‾ + 4 = 0 { |z| }^{ 2 }+(2+3i)z+(2-3i)\overline { z }+4=0

t = 2 − 3 i t=2-3i , k = 4 k=4

centre of this circle is − ( 2 − 3 i ) -(2-3i) , r a d i u s = t t ‾ − k radius=\sqrt{t \overline { t }-k}

∣ z + 2 − 3 i ∣ = 3 |z+2-3i|=3

let w = z + 3 + 2 i = 2 − 3 i + 1 + 5 i w=z+3+2i=2-3i+1+5i

∣ w − 1 − 5 i ∣ = ∣ z + 2 − 3 i ∣ = 3 |w-1-5i|=|z+2-3i|=3

so w w lies on circle whose centre is 1 + 5 i 1+5i and radius 3 3 units

7 Helpful 0 Interesting 0 Brilliant 0 Confused
Chew-Seong Cheong
May 23, 2015

Let z = x + y i z=x+yi , then:

∣ z 2 ∣ + 2 ( z + z ‾ ) + 3 i ( z − z ‾ ) + 4 = 0 x 2 + y 2 + 2 ( x + y i + x − y i ) + 3 i ( x + y i − x + y i ) + 4 = 0 x 2 + y 2 + 2 ( 2 x ) + 3 i ( 2 y i ) + 4 = 0 x 2 + y 2 + 4 x − 6 y + 4 = 0 ( x + 2 ) 2 − 4 + ( y − 3 ) 2 − 9 + 4 = 0 ( x + 2 ) 2 + ( y − 3 ) 2 = 9 \begin{aligned} |z^2|+2(z+\overline{z})+3i(z-\overline{z})+4 =0 \\ x^2+y^2+2(x+yi+x-yi) + 3i(x+yi-x+yi) +4 = 0 \\ x^2+y^2+2(2x) + 3i(2yi) +4 = 0 \\ x^2+y^2+4x -6y +4 = 0 \\ (x+2)^2-4 +(y-3)^2-9+4=0 \\ (x+2)^2+(y-3)^2=9 \end{aligned}

Therefore, z z lies on the circle with center − 2 + 3 i -2+3i and radius 3 3 . And z + 3 + 2 i z+3+2i has a center of ( − 2 + 3 ) + ( 3 + 2 ) i = 1 + 5 i (-2+3)+(3+2)i = 1+5i and radius 3 3 .

⇒ ∣ a + b + c ∣ = 1 + 5 + 3 = 9 \Rightarrow |a+b+c| = 1+5+3 = \boxed{9}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Converting into cartesian coordinates is one way to gain familiarity with the argand diagram.

Do you know how to describe a circle with center ω \omega and radius r r with a polar equation?

Thanks, for asking. Didn't about polar equation for a circle. Done some reading and it should be ρ = 2 a cos ⁡ θ + 2 b sin ⁡ θ \rho = 2a\cos{\theta} + 2b\sin{\theta} , where ω = a + b i \omega = a + bi .

Chew-Seong Cheong - 6 years ago

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