SOLUTION: I have 2 question i have answered both but am not confident with the answers.
(2+3i)(x+yi)=23+2i find values for x and y.
express root 2 cis (5pi/4) in cartesian form.
Thak you
Question 266199: I have 2 question i have answered both but am not confident with the answers.
(2+3i)(x+yi)=23+2i find values for x and y.
express root 2 cis (5pi/4) in cartesian form.
Thak you for your help. Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! (2+3i)(x+yi)=23+2i find values for x and y.
---
2x +2yi +3xi -3y = 23 + 2i
---
(2x-3y)+(3x+2y)i = 23 + 2i
---
Equate the Real Parts and Equate the Imaginary Parts to get:
2x-3y = 23
3x+2y = 2
------------------
Multiply 1st by 3; Multiply 2nd by 2
6x - 9y = 69
6x + 4y = 4
---------------
Subtract 1st from 2nd and solve for "y":
13y = -65
y = -5
-------------------
2x-3y = 23
3x+2y = 2
---
Multiply 1st by 2; Multiply 2nd by 3
4x -6y = 46
9x +6y = 6
-----------
Add the two equations and solve for "x":
13x = 52
x = 4
===========================================
express root 2 cis (5pi/4) in cartesian form.
---
= sqrt(2)[cos((5/4)pi)+ isin((5/4)pi))
= sqrt(2)[-1/sqrt(2) - i(1/sqrt(2)]
= [-1 -i]
==================
Cheers,
Stan H.
(