Can someone help me simplify this complex number?

Question

if z=3-4i
w= 1+3i
write z/w in polar form

Answer 1

(3 - 4i) / (1 + 3i)

Multiply by 1, which in this case will be (1 - 3i) / (1 - 3i) :

Note : (1 - 3i) is the conjugate of (1 + 3i). See how it all
comes together to make an integer for the denominator.

= (3 - 4i)(1 - 3i) / [(1 + 3i)(1 - 3i)]

Expand :

= (3 - 9i - 4i + 12i^2) / (1 - 9i^2)

= (3 - 9i - 4i - 12) / (1 + 9) because i^2 = -1

= (-9 - 13i) / 10

= (-9/10) + i(-13/10)
This is equivalent to x + iy, where x = -9/10 and y = -13/10.

The polar form is r[cos(t) + i*sin(t)]
where : r = sqrt(x^2 + y^2)
So, r = sqrt[(-9/10)^2 + (-13/10)^2] = sqrt(2.5)

and, t = arctan(y/x) - 180º (because x and y are both < 0)
Therefore, t = 55.3º - 180º = -124.7º
Adding 360º to make it a positive angle gives 235.3º

Polar form is : sqrt(2.5)[cos(235.3º) + i*sin(235.3º)]

Answer 2

(3-4i)/(1+3i)
so that we do not have an imaginary no. in the denominator, we now multiply both numerator and denominator by the complex conjugate of the denominator, which is 1-3i. We have

(3-4i)/(1+3i) * (1-3i)/(1-3i)
= (3 - 9i - 4i -12i^2) / (1 - 9i^2)
= (3 - 9i - 4i -12(-1)) / (1 - 9(-1))
= (3 - 9i - 4i +12) / (1 + 9)
= (15 - 13i) / 10
= 15/10 - (13/10)i
= 3/2 - (13/10)i

r = √[(3/2)^2 + (13/10)^2]
=√[9/4 + 169/100]
=√[(225 + 169)/100]
=√(394/100)
=√394 / 10

θ = Arctan[(-13/10)/(3/2)]
θ = Arctan[-13/15]
θ = -40.91 = 319.09deg

any complex number in z=x+iy form can be expressed in polar form z=r(sinθ + cosθ).
you may just plug in the values above...

Answer 3

That's a curly one. Let's see.

z/w = (3 - 4i)/(1 + 3i)

Multiplying top and bottom by 1 - 3i, we get:

z/w = (3 - 4i)(1 - 3i)/(1 - 9i^2)

= (4 - 12 - 13i)/10

= (- 8 - 13i)/10

= (-8/10) + (-13/10)i

From memory, the polar form of a complex number x + iy (where x and y are real numbers) is

r.e^(i.theta)

where r = sqrt(x^2 + y^2)

and theta = arctan (y/x)

So,

r = sqrt [ (-8/10)^2 + (-13/10)^2]

= sqrt [ (64/100) + (169/100)]

= sqrt (233/100)

= 1.5264 (approx.)

theta = arctan [(-13/10)/(-8/10)]

= arctan (13/8)

= arctan (1.625)

= 1.0191 (approx.)

So, the polar form of z/w is

sqrt(2.31).e^[i.arctan(1.625)]

or approximately

1.5264 e ^(1.0191 i)

It is about 30 years since I studied complex analysis, so I will be delighted if I haven't made any mistakes!

Answer 4

It no longer ordinary to form. So i will say the stairs. First go away the ^7 because it rather is. Simplify the interior parameter by ability of taking conjugate of the denominator and it is often (a million+sqrt(3))+(a million-sqrt(3))i = a + bi=z the flexibility 7 is extensive, so we are in a position to transform 'z' into polar style z=re^(i theta) r=sqrt(a^2 + b^2) theta=taninverse or in basic terms arctan(b/a) e^(i theta)=cos(theta) + i sin(theta) finally carry on with the De Moivres theorem [cos(theta) + i sin(theta)]^n= [cos n(theta) + i sin n(theta)] U'll get theta=(a million-sqrt(3))/(a million+sqrt(3)) n=7 and r=8. U can convert it returned to rect style yet no longer needed.

Answer 5

z/w = ( 3 - 4 i ) / ( 1 + 3 i )
z/w = ( 3 - 4 i ) ( 1 - 3 i ) / ( 1 ² + 3 ² )
z/w = ( 3 - 13 i - 12 ) / 10
z/w = ( - 9 - 13 i ) / 10
z/w = ( - 1 / 10 ) ( 9 + 13 i )
z/w = ( - 1 / 10 ) √250 /_55.3°
z/w = (- √10/10) (5) /_55.3°
z/w = (- √10/2) /_55.3°

Answer 6

(3-4i)/(1+3i)
(3-4i)/(1+3i) * (1-3i)/(1-3i)
(3 - 12 - 4i - 9i) / (1+9)
(-9 - 13i)/10
-0.9 - 1.3i
r = sqrt(0.9^2 + 1.3^2) = sqrt(2.5)
angle = pi/2 + arctan(1.3/0.9) = 2.5360 radians or 235.3048 degrees (approx.)
r cis(angle) = sqrt(2.5) cis(235.3048 degrees)

Answer 7

yes