Answer
a) $7+i$
b) $-1-5i$
c) $18-i$
d) ${6\over25}-{17\over25}i$
e) $1$
f) $2-2i$
Work Step by Step
a) we add the real and complex parts: $3+4+3i-2i=7+i$
b) we subtract the real and complex parts: $3-4 -2i-3i=-1-5i$
c)We apply the Binomial Multiplication:
$(3-2i)(4+3i)=$
$3(4)+3(3i)-2i(4)-2i(3i)=$
$12+9i-8i-6i^2$
Since $i^2=-1$
$12-i-6(-1)=$
$12-i+6=$
$18-i$
d) We multiply the numerator and denominator by the conjugate of the denominator:
${3-2i\over{4+3i}} \cdot{4-3i\over{4-3i}}=$
Now we apply the binomial multiplication
${{3(4)+3(-3i)-2i(4)-2i(-3i)}\over{4(4)+4(-3i)+3i(4)+3i(-3i)}}=$
${12-9i-8i+6i^2}\over{16-12i+12i-9i^2}$
Knowing that $i^2=-1$
${{12-17i+6(-1)}\over{16-9(-1)}}=$
${{12-17i-6}\over{16+9}}=$
${{6-17i}\over{25}}=$
${6\over25}-{17\over25}i$
e) We divide the exponent by 4 and the answer to the question will be $i$ to the power of the remainder of the division.
$48$ divided by $4$ gives $12$ with no remainder. So,
$i^{48}=i^0=1$
f)We apply the Binomial Multiplication:
$\sqrt2\sqrt8+\sqrt2\sqrt{-2}-\sqrt{-2}\sqrt8-\sqrt{-2}\sqrt{-2}=$
$\sqrt{16}+\sqrt{-4}-\sqrt{-16}-\sqrt4$
Knowing that $\sqrt{-n}=\sqrt n\cdot i$
$4+\sqrt4 \cdot i- \sqrt{16} \cdot i -2=$
$2+2i-4i=$
$2-2i$
