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        <title>MathCity.org</title>
        <description>Merging man &amp; maths</description>
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        <item>
            <title>Question 5, Exercise 1.3</title>
            <link>https://www.mathcity.org/fsc-part1-kpk/sol/unit01/ex1-3-p4</link>
            <description>Question 5, Exercise 1.3

Solutions of Question 5 of Exercise 1.3 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 5(i)
${{z}^{2}}+z+3=0$${{z}^{2}}+z+3=0$$a=1,\,\,\,b=1$$c=3$\begin{align}z&amp;=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}\\ 
z&amp;=\dfrac{-\left( 1 \right)\pm \sqrt{{{\left( 1 \right)}^{2}}-4\left( 1 \right)\left( 3 \right)}}{2\left( 1 \right)}\\
z&amp;=\dfrac{-1\pm \s…</description>
            <author>anonymous@undisclosed.example.com (Anonymous)</author>
            <pubDate>Mon, 25 Sep 2023 12:02:44 +0000</pubDate>
        </item>
        <item>
            <title>Question 6, Exercise 1.3</title>
            <link>https://www.mathcity.org/fsc-part1-kpk/sol/unit01/ex1-3-p5</link>
            <description>Question 6, Exercise 1.3

Solutions of Question 6 of Exercise 1.3 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 6(i)
${{z}^{4}}+{{z}^{2}}+1=0$\begin{align}{{z}^{4}}+{{z}^{2}}+1&amp;=0\\
{{z}^{4}}+2\left( \dfrac{1}{2} \right){{z}^{2}}+\dfrac{1}{4}-\dfrac{1}{4}+1&amp;=0\\
{{\left( {{z}^{2}}+\dfrac{1}{2} \right)}^{2}}+\dfrac{4-1}{4}&amp;=0\\
{{\left( {{z}^{2}}+\dfrac{1}{2} \righ…</description>
            <author>anonymous@undisclosed.example.com (Anonymous)</author>
            <pubDate>Mon, 25 Sep 2023 12:03:16 +0000</pubDate>
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        <item>
            <title>Question 1, Exercise 1.3</title>
            <link>https://www.mathcity.org/fsc-part1-kpk/sol/unit01/ex1-3-p1</link>
            <description>Question 1, Exercise 1.3

Solutions of Question 1 of Exercise 1.3 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 1(i)
\begin{align}&amp;z-4w=3i\\ 
&amp;2z+3w=11-5i\end{align}\begin{align}z-4w&amp;=3i		…(i)\\
2z+3w&amp;=11-5i	…(ii)\end{align}$2$\begin{align}2z-8w&amp;=6i		…(iii)\end{align}\[\begin{array}{cccc}
2z&amp;-8w&amp;=6i  \\  
\mathop+\limits_{-}2z&amp;\mathop+\limits_{-}3w&amp;=\mathop-\limit…</description>
            <author>anonymous@undisclosed.example.com (Anonymous)</author>
            <pubDate>Sat, 30 Sep 2023 18:43:14 +0000</pubDate>
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        <item>
            <title>Question 3 &amp; 4, Exercise 1.3</title>
            <link>https://www.mathcity.org/fsc-part1-kpk/sol/unit01/ex1-3-p3</link>
            <description>Question 3 &amp; 4, Exercise 1.3

Solutions of Question 3 &amp; 4 of Exercise 1.3 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Question 3
${{z}_{1}}=-1+i$${{z}_{2}}=-1-i$${{z}^{2}}+2z+2=0$$$z^2+2z_1+2=0\quad \ldots (i)$$$z_1=-1+i$\begin{align}L.H.S &amp;= (-1+i)^2+2(-1+i)+2\\
&amp;=1-2i-1-2+2i+2\\
&amp;=0=R.H.S\end{align}$z_1=-1+i$$z_2=-1-i$\begin{align}
L.H.S&amp;=(-1-i)^2+2(-1-i)+2\\
&amp;=1+2i-1-…</description>
            <author>anonymous@undisclosed.example.com (Anonymous)</author>
            <pubDate>Tue, 03 Oct 2023 03:15:42 +0000</pubDate>
        </item>
        <item>
            <title>Question 6, 7 &amp; 8, Review Exercise 1</title>
            <link>https://www.mathcity.org/fsc-part1-kpk/sol/unit01/review-ex-1-p4</link>
            <description>Question 6, 7 &amp; 8, Review Exercise 1

Solutions of Question 6, 7 &amp; 8 of Review Exercise 1 of Unit 01: Complex Numbers. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.$\dfrac{1}{3+4i}$\begin{align}\dfrac{1}{3+4i}&amp;=\dfrac{1}{3+4i}\times \dfrac{3-4i}{3-4i}\\
&amp;=\dfrac{3-4i}{9+16}\\
&amp;=\dfrac{3-4i}{25}\\
&amp;=\dfrac{3}{25}-\dfrac{4i}{25}\end{align}$\dfrac{3i+2}{3-2i}$\begin{align}\dfrac{3i+2}{3-2i}\\
\dfrac{3i+2}…</description>
            <author>anonymous@undisclosed.example.com (Anonymous)</author>
            <pubDate>Mon, 25 Sep 2023 12:10:41 +0000</pubDate>
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