# Solve for l z*z=(2+3l)(5-2i)

z⋅z=(2+3l)(5-2i)
Rewrite the equation as (2+3l)(5-2i)=z⋅z.
(2+3l)(5-2i)=z⋅z
Simplify.
Raise z to the power of 1.
(2+3l)(5-2i)=z1⋅z
Raise z to the power of 1.
(2+3l)(5-2i)=z1⋅z1
Use the power rule aman=am+n to combine exponents.
(2+3l)(5-2i)=z1+1
(2+3l)(5-2i)=z2
(2+3l)(5-2i)=z2
Divide each term by 5-2i and simplify.
Divide each term in (2+3l)(5-2i)=z2 by 5-2i.
(2+3l)(5-2i)5-2i=z25-2i
Cancel the common factor of 5-2i.
Cancel the common factor.
(2+3l)(5-2i)5-2i=z25-2i
Divide 2+3l by 1.
2+3l=z25-2i
2+3l=z25-2i
Simplify z25-2i.
Multiply the numerator and denominator of z25-2i by the conjugate of 5-2i to make the denominator real.
2+3l=z25-2i⋅5+2i5+2i
Multiply.
Combine.
2+3l=z2(5+2i)(5-2i)(5+2i)
Simplify the numerator.
Apply the distributive property.
2+3l=z2⋅5+z2(2i)(5-2i)(5+2i)
Move 5 to the left of z2.
2+3l=5⋅z2+z2(2i)(5-2i)(5+2i)
Move 2 to the left of z2.
2+3l=5z2+2z2i(5-2i)(5+2i)
2+3l=5z2+2z2i(5-2i)(5+2i)
Simplify the denominator.
Expand (5-2i)(5+2i) using the FOIL Method.
Apply the distributive property.
2+3l=5z2+2z2i5(5+2i)-2i(5+2i)
Apply the distributive property.
2+3l=5z2+2z2i5⋅5+5(2i)-2i(5+2i)
Apply the distributive property.
2+3l=5z2+2z2i5⋅5+5(2i)-2i⋅5-2i(2i)
2+3l=5z2+2z2i5⋅5+5(2i)-2i⋅5-2i(2i)
Simplify.
Multiply 5 by 5.
2+3l=5z2+2z2i25+5(2i)-2i⋅5-2i(2i)
Multiply 2 by 5.
2+3l=5z2+2z2i25+10i-2i⋅5-2i(2i)
Multiply 5 by -2.
2+3l=5z2+2z2i25+10i-10i-2i(2i)
Multiply 2 by -2.
2+3l=5z2+2z2i25+10i-10i-4ii
Raise i to the power of 1.
2+3l=5z2+2z2i25+10i-10i-4(i1i)
Raise i to the power of 1.
2+3l=5z2+2z2i25+10i-10i-4(i1i1)
Use the power rule aman=am+n to combine exponents.
2+3l=5z2+2z2i25+10i-10i-4i1+1
2+3l=5z2+2z2i25+10i-10i-4i2
Subtract 10i from 10i.
2+3l=5z2+2z2i25+0-4i2
2+3l=5z2+2z2i25-4i2
2+3l=5z2+2z2i25-4i2
Simplify each term.
Rewrite i2 as -1.
2+3l=5z2+2z2i25-4⋅-1
Multiply -4 by -1.
2+3l=5z2+2z2i25+4
2+3l=5z2+2z2i25+4
2+3l=5z2+2z2i29
2+3l=5z2+2z2i29
2+3l=5z2+2z2i29
Factor z2 out of 5z2+2z2i.
Factor z2 out of 5z2.
2+3l=z2⋅5+2z2i29
Factor z2 out of 2z2i.
2+3l=z2⋅5+z2(2i)29
Factor z2 out of z2⋅5+z2(2i).
2+3l=z2(5+2i)29
2+3l=z2(5+2i)29
2+3l=z2(5+2i)29
2+3l=z2(5+2i)29
Subtract 2 from both sides of the equation.
3l=z2(5+2i)29-2
Divide each term by 3 and simplify.
Divide each term in 3l=z2(5+2i)29-2 by 3.
3l3=z2(5+2i)29-23
Cancel the common factor of 3.
Cancel the common factor.
3l3=z2(5+2i)29-23
Divide l by 1.
l=z2(5+2i)29-23
l=z2(5+2i)29-23
Simplify z2(5+2i)29-23.
Simplify the numerator.
To write -2 as a fraction with a common denominator, multiply by 2929.
l=z2(5+2i)29-2⋅29293
Combine -2 and 2929.
l=z2(5+2i)29+-2⋅29293
Combine the numerators over the common denominator.
l=z2(5+2i)-2⋅29293
Rewrite z2(5+2i)-2⋅2929 in a factored form.
Apply the distributive property.
l=z2⋅5+z2(2i)-2⋅29293
Move 5 to the left of z2.
l=5⋅z2+z2(2i)-2⋅29293
Move 2 to the left of z2.
l=5z2+2z2i-2⋅29293
Multiply -2 by 29.
l=5z2+2z2i-58293
l=5z2+2z2i-58293
l=5z2+2z2i-58293
Multiply the numerator by the reciprocal of the denominator.
l=5z2+2z2i-5829⋅13
Multiply 5z2+2z2i-5829⋅13.
Multiply 5z2+2z2i-5829 and 13.
l=5z2+2z2i-5829⋅3
Multiply 29 by 3.
l=5z2+2z2i-5887
l=5z2+2z2i-5887
l=5z2+2z2i-5887
l=5z2+2z2i-5887
Solve for l z*z=(2+3l)(5-2i)

### Solving MATH problems

We can solve all math problems. Get help on the web or with our math app

Scroll to top