# Question 2, Exercise 1.1

Solutions of Question 2 of Exercise 1.1 of Unit 01: Complex Numbers. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Write the following complex number in the form $x+iy$: $(3+2i)+(2+4i)$

Solution.

\begin{align}&(3+i2)+(2+i4)\\ =&(3+2)+(i2+i4)\\ =&5+i6\end{align}

Write the following complex number in the form $x+iy$: $(4+3i)-(2+5i)$

Solution.

\begin{align}&(4+3i)-(2+5i)\\ =&(4-2)+(3i-5i)\\ =&2-2i\end{align}

Write the following complex number in the form $x+iy$: $(4+7i)+(4-7i)$

Solution.

\begin{align} &(4+7i)+(4-7i)\\ =&(4+4)+(7i-7i)\\ =&8+0i. \end{align}

Write the following complex number in the form $x+iy$: $(2+5i)-(2-5i)$

Solution.

\begin{align} &(2+5i)-(2-5i)\\ =&(2-2)+(5i+5i)\\ =&0+10i. \end{align}

Write the following complex number in the form $x+iy$: $(3+2i)(4-3i)$

Solution.

\begin{align}&(3+2i)(4-3i)\\ =&12-9i+8i-6i^2\\ =&12+6-9i+8i\\ =&18-i\end{align}

Write the following complex number in the form $x+iy$: $(3,2)\div(3,-1)$

Solution.

\begin{align}&(3,2)\div(3,-1)\\ =&\dfrac{3+2i}{3-i}\\ =&\dfrac{3+2i}{3-i}\times\dfrac{3+i}{3+i}\\ =&\dfrac{(3+2i)(3+i)}{3^2-i^2}\\ =&\dfrac{9+2i^2+6i+3i}{9+1}\\ =&\dfrac{9-2+9i}{10}\\ =&\dfrac{7+9i}{10}\\ =&\dfrac{7}{10}+\dfrac{9}{10}i\end{align}

Write the following complex number in the form $x+iy$: $(1+i)(1-i)(2+i)$

Solution.

\begin{align}&(1+i)(1-i)(2+i)\\ =&(1+i)(2-i^2+i-2i)\\ =&(1+i)(2+1-i)\\ =&(1+i)(3-i)\\ =&3-i^2+3i-i\\ =&4+2i\end{align}

Write the following complex number in the form $x+iy$: $\dfrac{1}{2+3i}$.

Solution.

\begin{align} &\dfrac{1}{2+3i} \\ =&\dfrac{1}{2+3i}\times\dfrac{2-3i}{2-3i} \\ =&\dfrac{2-3i}{4-9i^2}\\ =&\dfrac{2-3i}{4+9}\\ =&\dfrac{2}{13}-\dfrac{3}{13}i.\\ \end{align}