Loading [MathJax]/jax/output/SVG/config.js
 
+0  
 
0
3516
1
avatar

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

Write your solutions in a+bi form, separated by commas. So, "1+2i, 3-i" is an acceptable answer format, but "2i+1; -i+3" is not. (Don't include quotes in your answer.) 

Note: This problem is not about functions.

 Nov 6, 2014

Best Answer 

 #1
avatar+33658 
+7

Write z = (a + ib), then z2 = a2 - b2 +2abi

 

For this to equal 2i we must have

a2 - b2 = 0  and ab = 1

 

There are four solution sets for these: a = 1, b = 1; a = -1, b= -1; a = i, b=-i; a = -i, b = i.

 These give rise to just two different solutions:

1 + i  and -1 - i

.

 Nov 7, 2014
 #1
avatar+33658 
+7
Best Answer

Write z = (a + ib), then z2 = a2 - b2 +2abi

 

For this to equal 2i we must have

a2 - b2 = 0  and ab = 1

 

There are four solution sets for these: a = 1, b = 1; a = -1, b= -1; a = i, b=-i; a = -i, b = i.

 These give rise to just two different solutions:

1 + i  and -1 - i

.

Alan Nov 7, 2014

1 Online Users

avatar