Question

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

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

Answer 1

`"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`

Answer 2

`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`.

Answer 3

`\`

`\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.`