w and z are 2 different complex numbers.

|wz|=40

|w+z|=13

w = 3+4i

find z

doorknoob Nov 4, 2019

#1

#2**+1 **

**........ I mean what I said 4i**

**4i is a complex number....**

**if you you do not know what that is then you may as well not bother attempting this problem as you need an understanding of magnitude and complex numbers to solve it.**

doorknoob
Nov 5, 2019

#5**0 **

Hi doorknob,

Tomsun was trying to help you, he/she obviously did not know what complex numbers are.

That would be true of most people here.

Please do not be rude to people who take an interest in your questions.

You could have just said something like,

"Hi Tomsun,

4i is a complex number, you will probably learn about those in the future."

That would have sounded much nicer.

Melody
Nov 5, 2019

#3**+4 **

I haven't taken Alg 2 yet, but I can attempt it.

Imaginary number is \(\sqrt{-1}\)

So

\(w=3+\sqrt{-4}\)

Plugging in \(|w+z|=13\)

We have

\(3+\sqrt{-4}+z=13\)

\(\sqrt{-4}+z=10\)

We have

\(z=10-\sqrt{-4}\)

\(z=10-4i\)

Does this even work??

CalculatorUser Nov 5, 2019

#4**+5 **

**w and z are 2 different complex numbers.**

**|wz|=40**

**|w+z|=13**

**w = 3+4i**

**find z**

\(\text{Let $z=x+yi$}\)

**1.**

\(\begin{array}{|rcll|} \hline \mathbf{|wz|} &=& \mathbf{40} \\ \hline wz &=& (3+4i)(x+yi) \\ &=& 3x+3yi+4xi-4y \quad | \quad i^2 = -1 \\ &=& 3x-4y+(4x+3y)i \\ |wz| &=& \sqrt{(3x-4y)^2+(4x+3y)^2} \\ &=& \sqrt{ 9x^2-24xy+16y^2+16x^2+24xy+9y^2 } \\ &=& \sqrt{ 25x^2 +25y^2} \\ |wz| &=& 5\sqrt{ x^2 +y^2} \quad | \quad |wz| = 40 \\ 40 &=& 5\sqrt{ x^2 +y^2} \quad | \quad : 5 \\ 8 &=& \sqrt{ x^2 +y^2} \\ \sqrt{ x^2 +y^2} &=& 8 \\ \mathbf{x^2 +y^2} &=& \mathbf{64} \qquad (1) \\ \hline \end{array}\)

**2.**

\(\begin{array}{|rcll|} \hline \mathbf{|w+z|} &=& \mathbf{13} \\ \hline w + z &=& (3+4i)+(x+yi) \\ &=& (3+x) + (4+y)i \\ |w+z| &=& \sqrt{(3+x)^2+(4+y)^2 } \\ &=& \sqrt{ 9+6x+x^2+16+8y+y^2 } \\ &=& \sqrt{ 25+6x+8y +x^2+y^2 } \quad | \quad x^2+y^2 = 64 \\ &=& \sqrt{ 25+6x+8y + 64 } \\ |w+z| &=& \sqrt{ 89+6x+8y } \quad | \quad |w+z| = 13 \\ 13 &=& \sqrt{ 89+6x+8y } \\ 13^2 &=& 89+6x+8y \quad | \quad -89 \\ 80 &=&6x+8y \quad | \quad : 2 \\ 40 &=& 3x+4y \\ 4y &=& 40-3x \\ \mathbf{y} &=& \mathbf{\dfrac{40-3x }{4}} \qquad (2) \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{x^2 +y^2} &=& \mathbf{64} \quad | \quad \mathbf{y=\dfrac{40-3x }{4}} \\\\ x^2+\left( \dfrac{40-3x }{4} \right)^2 &=& 64 \\ x^2+ \dfrac{(40-3x)^2 }{16} &=& 64 \quad | \quad \cdot 16 \\ 16x^2+ (40-3x)^2 &=& 1024 \\ 16x^2+40^2-240x+9x^2 &=& 1024 \\ 25x^2-240x +576 &=& 0 \\\\ x &=& \dfrac{240 \pm \sqrt{240^2-4\cdot 25\cdot 576} }{2\cdot 25 } \\ x &=& \dfrac{240 \pm \sqrt{57600 -57600} }{50} \\ x &=& \dfrac{240}{50} \\ \mathbf{x} &=& \mathbf{\dfrac{24 }{5} } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{y} &=& \mathbf{\dfrac{40-3x }{4}} \quad | \quad \mathbf{x=\dfrac{24}{5}} \\\\ y &=& \dfrac{40-3\cdot \dfrac{24}{5} }{4} \\ y &=& 10-3\cdot \dfrac{6}{5} \\ y &=& \dfrac{50-18}{5} \\ y &=& \dfrac{32}{5} \\ \mathbf{y} &=& \mathbf{\dfrac{32}{5}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{z} &=& \mathbf{\dfrac{24}{5} + \dfrac{32}{5}i } \\ \hline \end{array}\)

heureka Nov 5, 2019