SOLUTION: Hi, can someone please help, . Imaginary Numbers . Simplify {{{(5-3i)/(-3-2i) }}} . Put in complex number form {{{ a + bi }}} . Thanks ahead of time, Levi

Algebra ->  Real-numbers -> SOLUTION: Hi, can someone please help, . Imaginary Numbers . Simplify {{{(5-3i)/(-3-2i) }}} . Put in complex number form {{{ a + bi }}} . Thanks ahead of time, Levi      Log On


   

Question 159163: Hi, can someone please help,
.
Imaginary Numbers
.
Simplify %285-3i%29%2F%28-3-2i%29+
.
Put in complex number form +a+%2B+bi+
.
Thanks ahead of time, Levi
Found 2 solutions by nerdybill, ankor@dixie-net.com:
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
+%285-3i%29%2F%28-3-2i%29+
.
To get rid of the 'i' from the numerator, multiply by (-3+2i)/(-3+2i):
+%285-3i%29%2F%28-3-2i%29+%2A+%28-3%2B2i%29%2F%28-3%2B2i%29+
+%28%285-3i%29%28-3%2B2i%29%29%2F%289-4%29+
+%28%285-3i%29%28-3%2B2i%29%29%2F%285%29+
+%28-15%2B10i%2B9i%2B6%29%2F%285%29+
+%28-9%2B19i%29%2F%285%29+
+-9%2F5+%2B+%2819%2F5%29i+

Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
Multiply by the conjugate of the denominator over itself:
%28%285-3i%29%29%2F%28%28-3-2i%29%29 * %28%28-3%2B2i%29%29%2F%28%28-3%2B2i%29%29 = %28%28-15%2B10i%2B9i-6%28i%5E2%29%29%29%2F%28%289+-+4%28i%5E2%29%29%29 = %28%28-15%2B10i%2B9i-6%28-1%29%29%29%2F%28%289+-+4%28-1%29%29%29 =
%28%28-15%2B10i%2B9i%2B6%29%29%2F%28%289%2B4%29%29 = %28%28-9%2B19i%29%29%2F13