How do I calculate the Real and Imaginary Parts of this equation?

#z=(e^(2+ipi/2)/(1+3i))^2#

3 Answers
Feb 11, 2018

#"Real part "= 0.08*e^4#
#"and Imaginary part "= 0.06*e^4#

Explanation:

#exp(a+b) = e^(a+b) = e^a * e^b = exp(a)*exp(b)#
#exp(i theta) = cos(theta) + i sin(theta)#
#=> e^(2+i*pi/2) = e^2*exp(i*pi/2) = e^2*(cos(pi/2)+i sin(pi/2))#
#= e^2 * (0 + i) = e^2 * i#

#1/(1 + 3i) = (1-3i)/((1-3i)(1+3i)) = (1-3i)/10 = 0.1 - 0.3 i#

#"So we have"#

#(e^2*i*(0.1-0.3 i))^2#
#= e^4*(-1)*(0.1-0.3*i)^2#
#= - e^4 * (0.01+0.09 * i^2 - 2*0.1*0.3* i)#
#= - e^4 * (-0.08 - 0.06*i)#
#= e^4 (0.08 + 0.06*i)#

#=> "Real part "= 0.08*e^4#
#"and Imaginary part "= 0.06*e^4#

Feb 11, 2018

# Rl(z)=2/25e^4, and, Im(z)=3/50e^4#.

Explanation:

Recall that, #e^(itheta)=costheta+isintheta..............(square)#.

#:. z=((e^(2+ipi/2))/(1+3i))^2#,

#=(e^(2+ipi/2))^2/(1+3i)^2#,

#=e^(2*(2+ipi/2))/(1+3i)^2#,

#=e^(4+ipi)/(1+3i)^2#,

#=(e^4*e^(ipi))/(1+3i)^2#,

#={e^4*(cospi+isinpi)}/(1+3i)^2#,

#={e^4(-1+i*0)}/(1+3i)^2#,

#=-e^4*1/(1+3i)^2*(1-3i)^2/(1-3i)^2#,

#=-{e^4(1-3i)^2}/{(1+3i)(1-3i)}^2#,

#=-{e^4(1-3i)^2}/(1-9i^2)^2#,

#=-(e^4(1-6i+9i^2))/{1-9(-1)}^2#,

#=-(e^4(1-6i-9))/(10)^2#,

#=-(e^4(-8-6i))/100#,

#=(e^4(4+3i))/50#.

#rArr Rl(z)=2/25e^4, and Im(z)=3/50e^4#.

Feb 11, 2018

# \ #

# \qquad \qquad \qquad \qquad \qquad \quad \ \ ( { e^{ 2 + i \pi/2 } } / { 1 + 3 i } )^2 \ = \ { 2 e^4 }/25 + { 3 e^4 }/50 i. #

Explanation:

# \ #

# "We will work this out, working on the complex exponential" #
# "part first." #

# "Here we go: " #

# ( { e^{ 2 + i \pi/2 } } / { 1 + 3 i } )^2 \ = \ ( e^{ 2 + i \pi/2 } )^2 / ( 1 + 3 i )^2 \ = \ ( e^{4 + i \pi } ) / ( 1 + 3 i )^2 \ = \ ( e^{ 4 } e^ {i \pi } ) / ( 1 + 3 i )^2 #

# \qquad \qquad \qquad = \ { e^{ 4 } ( cos( \pi ) + i sin( \pi ) ) } / ( 1 + 3 i )^2 \ = \ ( e^{ 4 } ( -1 + i \cdot 0 ) ) / ( 1 + 3 i )^2 #

# \qquad \qquad \qquad = \ e^4 \cdot {-1} / ( 1 + 3 i )^2 \ = \ e^4 \cdot {-1} / ( 1 + 3 i )^2 \cdot ( 1 - 3 i )^2 / ( 1 -3 i )^2 #

# \qquad \qquad \qquad = \ e^4 \cdot { -1 \cdot ( 1 - 3 i )^2 } / { ( 1 + 3 i )^2 ( 1 -3 i )^2 } \ = \ e^4 \cdot { -1 \cdot ( 1 - 3 i )^2 } / { [ ( 1 + 3 i ) ( 1 -3 i ) ]^2 } #

# \qquad \qquad \qquad = \ e^4 \cdot { -1 \cdot ( 1 - 6 i + 9 i^2 ) } / ( 1^2 + 3^2 )^2 \ = \ e^4 \cdot { -1 \cdot ( 1 - 6 i - 9 ) } / 10^2 #

# \qquad \qquad \qquad = \ e^4 \cdot { -1 \cdot ( -8 - 6 i ) } / 100 = \ e^4 \cdot { 8 + 6 i } / 100 #

# \qquad \qquad \qquad = \ e^4 cdot color(red)cancel{2} cdot ( 4 +3 i ) / { color(red)cancel{2} \cdot 50 } \ = \ e^4 cdot ( 4/50 +3/50 i ) #

# \qquad \qquad \qquad = \ e^4 cdot ( 2/25 +3/50 i ) \ = \ { 2 e^4 }/25 + { 3 e^4 }/50 i. #

# \ #

# "Thus:" #

# \qquad \qquad \qquad \qquad \qquad \qquad \ \ ( { e^{ 2 + i \pi/2 } } / { 1 + 3 i } )^2 \ = \ { 2 e^4 }/25 + { 3 e^4 }/50 i. #