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Home > Mathematics Homework Help > Solution

In step 1 why did you multiple Z by 2-3i/ 2-3i

Asked on March 27, 2026|Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

ScanSolve AI AnswerAnswered on March 27, 2026
To express a complex number in the form $x+yi$, the denominator must be a real number. Multiplying a complex number by its conjugate results in a real number: $$(a+bi)(a-bi) = a^2 - (bi)^2 = a^2 - b^2i^2 = a^2 + b^2$$ In Step 1, the denominator was $2+3i$. Its conjugate is $2-3i$. Multiplying the denominator by $2-3i$ makes it a real number: $$(2+3i)(2-3i) = 2^2 + 3^2 = 4+9 = 13$$ To maintain the value of the fraction, the numerator must also be multiplied by the same term, $2-3i$. This is equivalent to multiplying the entire fraction by $\frac{2-3i}{2-3i}$, which is equal to $1$.

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Home›Mathematics Homework Help›In step 1 why did you multiple Z by 2-3i/ 2-3i
Q

In step 1 why did you multiple Z by 2-3i/ 2-3i

March 27, 2026|Mathematics
ScanSolve AI

ScanSolve AI Answer

Answered on March 27, 2026

To express a complex number in the form x+yix+yix+yi, the denominator must be a real number.

Multiplying a complex number by its conjugate results in a real number: (a+bi)(a−bi)=a2−(bi)2=a2−b2i2=a2+b2(a+bi)(a-bi) = a^2 - (bi)^2 = a^2 - b^2i^2 = a^2 + b^2(a+bi)(a−bi)=a2−(bi)2=a2−b2i2=a2+b2 In Step 1, the denominator was 2+3i2+3i2+3i. Its conjugate is 2−3i2-3i2−3i. Multiplying the denominator by 2−3i2-3i2−3i makes it a real number: (2+3i)(2−3i)=22+32=4+9=13(2+3i)(2-3i) = 2^2 + 3^2 = 4+9 = 13(2+3i)(2−3i)=22+32=4+9=13 To maintain the value of the fraction, the numerator must also be multiplied by the same term, 2−3i2-3i2−3i. This is equivalent to multiplying the entire fraction by 2−3i2−3i\frac{2-3i}{2-3i}2−3i2−3i​, which is equal to 111.

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