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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

Best Answer 

 #8
avatar
+5

Thanks a lot!

You are a life saver 

Guest Jan 30, 2015
 #1
avatar
+5

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

Guest Jan 29, 2015
 #3
avatar+87301 
+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
avatar+19638 
+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.

   $$\\\small{\text{\ $z = -3$}}\\
\small{\text{ $a=-3, \ b = 0 $}}\\
\small{\text{ $r=\sqrt{(-3)^2+0^2} = 3$}} \\
\small{\text{ $a < 0, \ b=0:\quad$ $\phi=\arctan{( \frac{b}{a} )} + \pi
=\arctan{( \frac{0}{-3} )}+\pi = 0+\pi = \pi
$}}\\
\small{\text{$\textcolor[rgb]{1,0,0}{z = -3 = 3e^{i\pi}}$}}$$

 

II.

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

 

III.

$$\\\small{\text{\ $z = -5i$}}\\
\small{\text{ $a=0, \ b = -5$}}\\
\small{\text{ $r=\sqrt{(0)^2+(-5)^2} = 5$}} \\
\small{\text{ $a = 0, \ b < 0:\quad$ $\phi= -\frac{ \pi }{2} $}}\\
\small{\text{$\textcolor[rgb]{1,0,0}{z = -5i = 5e^{-\frac{ \pi }{2} i}}$}}$$

 

heureka  Jan 29, 2015
 #5
avatar
+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
avatar+26750 
+5

The last one is correct.

.

Alan  Jan 29, 2015
 #7
avatar+19638 
+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
avatar
+5
Best Answer

Thanks a lot!

You are a life saver 

Guest Jan 30, 2015

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