Question

Express the follwoing in the form a +bi

Answer 1

$\left(i\right)(8+7i)(8-7i)=8^2-(7i{)}^2=64-49i^2$
$=64-49\times -1=64+49=113=113+0i.\left(ii\right)(5-6i{)}^2=5^2-2(5\left)\right(6i)+(6i{)}^2=25-60i+36i^2=25-60i-36=-11-60i.\left(iii\right)(2+3i)(3+7i)=6+14i9i+21i^2=6+23i-21=-15+23i.$
$\left(iv\right){(-2-\frac{1}{3}i)}^3=(-1{)}^3{(2+\frac{1}{3}i)}^3=-(2^3+3\times 2^2\times \frac{1}{3}i+3\times 2\times {\left(\frac{1}{3}i\right)}^2+{\left(\frac{1}{3}i\right)}^3)$
$=-(8+4i+\frac{2}{3}{i}^2+\frac{1}{27}{i}^3)=-(8+4i+\frac{2}{3}(-1)+\frac{1}{27}(-i\left)\right)=-(8-\frac{2}{3}+4i-\frac{1}{27}i)=-(\frac{22}{3}+\frac{107}{27}i)=-\frac{22}{3}-\frac{107}{27}i.$
$\left(v\right)(1-l{)}^4=[(1-1{)}^2{]}^2=(1-2i+i^2{)}^2=(1-2i-1{)}^2(-2i{)}^2=4i^2=4\times -1=-4.$
$\left(vi\right)(\sqrt{3}+5i)(\sqrt{3}-5i{)}^2+(-4+5i{)}^2$
$=\left[\right(\sqrt{3}+5i\left)\right(\sqrt{3}-5i\left)\right](\sqrt{3}-5i)+(-4+5i{)}^2=(3-25i62\left)\right(\sqrt{3}-5i)+(16-40i+25i^2)=(3-25\times -1\left)\right(\sqrt{3}-5i)+(16-40i-25)=28(\sqrt{3}-5i)-9-40i=28\sqrt{3}-140i-9-40i=(28\sqrt{3}-9)-180i.$