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# Complex numbers into 2 different polar forms .. HELP!

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Hi. I just did these question but I am very unsecure about the answers. I am pretty sure that its wrong but I dont know why exactly. Please help!

Rewrite to polar form re ^ iθ, where θ is an angle in radians

• -3
• r = lzl  = √(a^2 + b^2) = √-3^2 = √9 = 3
• θ = tan-1(b/a) = tan-1(0/-3) = 0
• Result = 3*e^0i
•  1-i
• r = lzl  = √(a^2 + b^2) = √1^2+(-i)^2 = √1+1 = √2 = 1,41
• θ = tan-1(b/a) = tan-1(-1/1) = -pi/4
• Result = 1,41*e^i(-pi/4)
• -5i
• r = lzl  = √(a^2 + b^2) = √-5^2 = √25 = 5
• θ = tan-1(b/a) = tan-1(-5/0) = pi/2
• Result = 5*e^ipi/2

Rewrite the complex number to polar form r<θ, where θ is in degrees

• conjugate(4+2i)
• 4-2i
• r = lzl  = √(a^2 + b^2) = √4^2+(-2i)^2 = √4^2 + 2^2 = √16+4 = √20 = 4,47
• θ = tan-1(b/a) = tan-1(-2/4) = -26,5 degrees
• Result: 4,47 <-26,5 degrees
Guest Jan 29, 2015

#8
+5

Thanks a lot!

You are a life saver

Guest Jan 30, 2015
#1
+5

You can verify your answers easily with calculator: e^ia = cos(a) + i*sin(a)..

Guest Jan 29, 2015
#3
+90023
+5

I'm unfamiliar with the notation used in the last one, so I can't comment on that one

On the -5i one....think about where the terminal point of this angle lies.....it's on the y axis, 5 units "down" from the origin in the "imaginary" direction {pardon the non-math terms, here}

So, the correct angle should be -pi.2, not pi/2......

CPhill  Jan 29, 2015
#4
+20025
+5

Rewrite to polar form re ^ iθ, where θ is an angle in radians

$$\small{\text{\boxed{z=a+bi \quad \to \quad re^{i\phi} }}}\\\\ \small{\text{  \boxed{r=\sqrt{a^2+b^2} \qquad \phi= \left\{ \begin{array}{ll} \arctan{( \frac{b}{a} )}, & a > 0 \\ \\ \arctan{( \frac{b}{a} )} + \pi , & a < 0 \text{ and } b \ge 0 \\\\ \arctan{( \frac{b}{a} )} - \pi , & a < 0 \text{ and } b < 0 \\\\  \frac{ \pi }{2}, & a = 0 \text{ and } b > 0 \\\\ -\frac{ \pi }{2}, & a = 0 \text{ and } b < 0 \\\\ \end{array} \right. }  }}$$

I.

$$\\\small{\text{\ z = -3}}\\ \small{\text{ a=-3, \ b = 0 }}\\ \small{\text{ r=\sqrt{(-3)^2+0^2} = 3}} \\ \small{\text{ a < 0, \ b=0:\quad \phi=\arctan{( \frac{b}{a} )} + \pi =\arctan{( \frac{0}{-3} )}+\pi = 0+\pi = \pi }}\\ \small{\text{{z = -3 = 3e^{i\pi}}}}$$

II.

$$\\\small{\text{\ z = 1-i}}\\ \small{\text{ a=1, \ b = -1}}\\ \small{\text{ r=\sqrt{(1)^2+(-1)^2} = \sqrt{2}}} \\ \small{\text{ a > 0, \ b < 0:\quad \phi= \arctan{( \frac{b}{a} )} =\arctan{( \frac{-1}{1} )} = -\frac{\pi}{4} }}\\ \small{\text{{z = 1-i = \sqrt{2}e^{-\frac{\pi}{4} i}}}}$$

III.

$$\\\small{\text{\ z = -5i}}\\ \small{\text{ a=0, \ b = -5}}\\ \small{\text{ r=\sqrt{(0)^2+(-5)^2} = 5}} \\ \small{\text{ a = 0, \ b < 0:\quad \phi= -\frac{ \pi }{2} }}\\ \small{\text{{z = -5i = 5e^{-\frac{ \pi }{2} i}}}}$$

heureka  Jan 29, 2015
#5
+5

Thanks everyone!

Now its just the last one that I cant figure out if its right or wrong:

z = a+bi --> r and θ (degrees)

z = conjugate(4+2i) = 4-2i

r = lzl  = √(a^2 + b^2) = √4^2+(-2i)^2 = √4^2+2^2 = √16+4 = √20 = 4,47

b < 0 and a > 0

θ = tan-1(b/a) + pi = tan-1(-2/4) + pi = -23,42 degrees

4,47 and -23,42 degrees

Guest Jan 29, 2015
#6
+27052
+5

The last one is correct.

.

Alan  Jan 29, 2015
#7
+20025
+5

Now its just the last one that I cant figure out if its right or wrong:

z = a+bi --> r and θ (degrees)

z = conjugate(4+2i) = 4-2i

$$\small{\text{\boxed{r=\sqrt{a^2+b^2}\qquad\phi= \left\{ \begin{array}{ll}\arctan{( \frac{b}{a} )}, & a > 0 \text{ and } b \text{ whatever!} \\ \\\arctan{( \frac{b}{a} )} + \pi , & a < 0 \text{ and } b \ge 0 \\\\\arctan{( \frac{b}{a} )} - \pi , & a < 0 \text{ and } b < 0 \\\\\frac{ \pi }{2}, & a = 0 \text{ and } b > 0 \\\\-\frac{ \pi }{2}, & a = 0 \text{ and } b < 0 \\\\\end{array}\right.}}}$$

$$\\\small{\text{\ z = 4-2i}}\\ \small{\text{ a=4, \ b = -2}}\\ \small{\text{ r=\sqrt{(4)^2+(-2)^2} = \sqrt{20}=4.47}} \\ \small{\text{ a > 0 :\quad \phi= \arctan{( \frac{b}{a} )} =\arctan{( \frac{-2}{4} )} = -26.5650511771\ensurement{^{\circ}} }}$$

heureka  Jan 29, 2015
#8
+5