+0  
 
0
934
2
avatar+29 

Give a complex number z , so ;

a) |z|=π,z≠±π

b) e^z=e^(-1); z ≠ -1

c) ³√z=1+ i

 Oct 24, 2014

Best Answer 

 #2
avatar+23245 
+5

For the third problem:

Use deMoivre's Theorem:

Write the  complex number in r[cos(θ) +i·sin(θ)] form:          r = √(x²+y²)      θ = invTan(y/x)

z  =  1 + i   --->  x = 1 and y = 1         r = √(1²+1²)    --->   r  =  √2                  θ = invTan(1/1)   = 45°   

z^(1/3)  --->   r^(1/3)[cos(θ/3) + ·isin(θ/3)  --->  (√2)^(1/3)[cos(45°/3) +i·sin(45°/3)]

     --->  (2)^(1/6)[cos(15) + isin(15)]

then, add two more values by adding 360°/3  =  120°  and 2·360°/3 =  240°

--->    (2)^(1/6)[cos(135°) + isin(135°)]   and    (2)^(1/6)[cos(255°) + isin(255°)]

 Oct 24, 2014
 #1
avatar+23245 
+5

For the first problem:  The absolute value of a complex number is the distance from the origin to that point.

So, if you draw a circle whose center is the origin and whose radius is π, every point on that circle will solve the problem.

The two obvious answers are at x = π and x = -π, but you aren't allowed to use them; so choose any other location on the circle.

If you choose the angle 45°, θ= 45° and r = π,

the value of   z = r[cos(θ) + i ·sin(θ)]  becomes  π[cos(45°) + i · sin(45°)]  =  √2/2·π + √2/2·π·i

 Oct 24, 2014
 #2
avatar+23245 
+5
Best Answer

For the third problem:

Use deMoivre's Theorem:

Write the  complex number in r[cos(θ) +i·sin(θ)] form:          r = √(x²+y²)      θ = invTan(y/x)

z  =  1 + i   --->  x = 1 and y = 1         r = √(1²+1²)    --->   r  =  √2                  θ = invTan(1/1)   = 45°   

z^(1/3)  --->   r^(1/3)[cos(θ/3) + ·isin(θ/3)  --->  (√2)^(1/3)[cos(45°/3) +i·sin(45°/3)]

     --->  (2)^(1/6)[cos(15) + isin(15)]

then, add two more values by adding 360°/3  =  120°  and 2·360°/3 =  240°

--->    (2)^(1/6)[cos(135°) + isin(135°)]   and    (2)^(1/6)[cos(255°) + isin(255°)]

geno3141 Oct 24, 2014

4 Online Users

avatar
avatar
avatar