Question 151572
{{{ (9-8i)+(7+2i) }}} Start with the given expression.



{{{ (9+7)+(-8i+2i) }}} Group like terms



{{{ 16-6i }}} Combine like terms


So {{{(9-8i)+(7+2i)=16-6i}}}


So the expression is in standard form {{{a+bi}}} where {{{a=16}}} and {{{b=-6}}}


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{{{(3+4i)(2-6i)}}} Start with the given expression.



Now let's FOIL the expression.



Remember, when you FOIL an expression, you follow this procedure:



{{{(highlight(3)+4i)(highlight(2)-6i)}}} Multiply the <font color="red">F</font>irst terms:{{{(3)*(2)=6}}}.



{{{(highlight(3)+4i)(2+highlight(-6i))}}} Multiply the <font color="red">O</font>uter terms:{{{(3)*(-6*i)=-18*i}}}.



{{{(3+highlight(4i))(highlight(2)-6i)}}} Multiply the <font color="red">I</font>nner terms:{{{(4*i)*(2)=8*i}}}.



{{{(3+highlight(4i))(2+highlight(-6i))}}} Multiply the <font color="red">L</font>ast terms:{{{(4*i)*(-6*i)=-24*i^2=-24(-1)=24}}}.



{{{6-18*i+8*i+24}}} Now collect every term to make a single expression.



{{{30-10*i}}} Now combine like terms.



So {{{(3+4i)(2-6i)}}} FOILS to {{{30-10*i}}}.



In other words, {{{(3+4i)(2-6i)=30-10*i}}}.


So the expression is in standard form {{{a+bi}}} where {{{a=30}}} and {{{b=-10}}}



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{{{(6+3i)/(2+2i)}}} Start with the given expression



{{{((6+3i)/(2+2i))((2-2i)/(2-2i))}}} Multiply by the fraction by {{{(2-2i)/(2-2i)}}}. Note  {{{2-2i}}} is the conjugate of {{{2+2i}}}.



{{{((6+3i)(2-2i))/((2+2i)(2-2i))}}} Combine the fractions



{{{((6+3i)(2-2i))/(4+4i-4i-4i^2)}}} Foil the denominator



{{{((6+3i)(2-2i))/(4-4i^2)}}} Combine like terms



{{{((6+3i)(2-2i))/(4-4(-1))}}} Rewrite {{{i^2}}} as {{{-1}}}



{{{((6+3i)(2-2i))/(4+4)}}} Multiply



{{{((6+3i)(2-2i))/(8)}}} Add



{{{(12-12i+6i-6i^2)/(8)}}} FOIL the numerator



{{{(12-12i+6i-6(-1))/(8)}}} Rewrite {{{i^2}}} as {{{-1}}}



{{{(12-12i+6i+6)/(8)}}} Multiply



{{{(18-6i)/(8)}}} Combine like terms.



{{{(18)/(8)-(6i)/(8)}}} Break up the fraction



{{{(9)/(4)-(3/4)i}}} Reduce



So {{{(6+3i)/(2+2i)=(9)/(4)-(3/4)i}}}. 



So the expression is in standard form {{{a+bi}}} where {{{a=9/4}}} and {{{b=3/4}}}