Question

4i8-3i9+3÷3i11-4i10-2 express in a+bi form​

Answer 1

Answer:  The given expression is equal to $\dfrac{23}{13}+\dfrac{15}{13}i.$

Step-by-step explanation:  We are given to express the following complex expression in a + bi form :

$E=(4i^8-3i^9+3)\div(3i^{11}-4i^{10}-2)~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

We will be using the following values of the imaginary number i :

$i^2=-1,\\\\~~i^8=(-1)^4=1\\\\,~i^9=1\times i=i,\\\\~i^{10}=1\times(-1)=-1,\\\\~i^{11}=1\times(-1)\times i=-i.$

Therefore, from expression (i), we get

$E\\\\=(4i^8-3i^9+3)\div(3i^{11}-4i^{10}-2)\\\\\\=\dfrac{4i^8-3i^9+3}{3i^{11}-4i^{10}-2}\\\\\\=\dfrac{4\times1-3\times i+3}{3\times(-i)-4\times(-1)-2}\\\\\\=\dfrac{7-3i}{2-3i}\\\\\\=\dfrac{(7-3i)(2+3i)}{(2-3i)(2+3i)}\\\\\\=\dfrac{14+21i-6i-9i^2}{4+9}\\\\\\=\dfrac{23+15i}{13}\\\\\\=\dfrac{23}{13}+\dfrac{15}{13}i.$

Thus, the given expression is equal to $\dfrac{23}{13}+\dfrac{15}{13}i.$