+0  
 
+2
216
4
avatar+415 

Suppose z is a complex number such that \(z^2=24-32i\). Find \(|z|\).

 Mar 11, 2021
 #1
avatar+524 
+2

z = 24 -32i

Thus, 

 

|z| = \({\sqrt{24^2 + 32^2}}\)

 

     = \({\sqrt{1600}}\)

 

     = 40

 Mar 11, 2021
edited by amygdaleon305  Mar 11, 2021
 #2
avatar+415 
0

Question:

Wouldn't it be \(\sqrt40\) or \(2\sqrt10\) ?

MobiusLoops  Mar 11, 2021
edited by MobiusLoops  Mar 11, 2021
 #4
avatar+524 
+1

No, this answer was wrong. I've posted the correct one.

amygdaleon305  Mar 11, 2021
 #3
avatar+524 
+3

Turns out, the previous answer I posted was wrong.

Here's the correct solution :- 

 

z2 = 24 - 32i

 

Let z = x + iy

⇒ z2 = x+ i2y2 + 2xiy

         = x2 - y2 + 2xiy

 

Now, the real part,

x2 - y2 = 24                            ...(1)

 

Imaginary part, 

2xy = -32

xy   = -16

x     = -16/y

 

From (1)

x2 - 256/x2 = 24

x- 256 = 24x2

x4 - 24x2 - 256 = 0 

 

Solving this equation, 

x2 = -8, 36

⇒x = 6 

   y = -8/3

 

Thus z = 6 - 8i/3 

         |z| = \({ \sqrt{6^2 + (8/3)^2}}\)

              = 6.57

 

 

Hope this helps :)

 Mar 11, 2021

14 Online Users

avatar